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2.4 Conditional Probability

A generic candlestick chart.

2.4: Conditional Probability

Learning Objectives

Upon completion of this section, you should be able to

  • Compute conditional probabilities
  • Compute probability using the multiplication rule
  • Compute probability with independent events

Conditional Probability

Imagine you’re playing a game with a fair six-sided die. Someone rolls the die behind a screen and asks you, “What’s the probability that it’s a five?” Initially, you’d say 16, right? There are six possible outcomes, and they’re all equally likely.

Now imagine if they then tell you, “The number is odd”? This new information changes things. Now you know the roll must be 1, 3, or 5. Out of these three possibilities, only one is a five. So your new answer would be 13.

This scenario introduces us to conditional probability. Conditional probability asks: “What is the likelihood of an event, given that we know something else has already happened?” In probability notation, we write this as P(A|B), which means “the probability of event A, given that event B has occurred.”

In our die example:
Event A: Rolling a five
Event B: Rolling an odd number
P(A|B): The probability of rolling a five, given that we know the roll was odd

This concept of updating probabilities based on new information is crucial in many real-world applications, from weather forecasting to medical diagnoses.

Conditional Probability

Conditional probability is the likelihood of an event occurring, given that another event has already occurred. It measures how the probability of one event changes when we have information about another related event.

The conditional probability of event A occurring, given that event B has occurred, is denoted as P(A|B) and is read as “the probability of A given B.”

General Formula:
PA|B=PABPB

Where PB0

Example 1

A fair die is rolled.

  1. Find the probability that the number rolled is a five, given that it is odd.
  2. Find the probability that the number rolled is odd, given that it is a five.

Solution

The sample space for this experiment is the set S={1,2,3,4,5,6} consisting of six equally likely outcomes. Let F denote the event “a five is rolled” and let O denote the event “an odd number is rolled,” so that F={5} and O={1,3,5}.

  1. This is the introductory example, so we already know that the answer is 13. To use the formula in the definition to confirm this we must replace A in the formula (the event whose likelihood we seek to estimate) by F and replace B (the event we know for certain has occurred) by O:

    P(F|O)=P(FO)P(O)

    We know FO={5}, so this tell us that P(FO)=16

    We also know that O={1,3,5}, which tells us that P(O)=36

    Thus,

    P(F|O)=P(FO)P(O)=1636=1663=13

  2. This is the same problem, but with the roles of F and O reversed. Since we are given that the number that was rolled is five, which is odd, the probability in question must be 1. To apply the formula to this case we must now replace A (the event whose likelihood we seek to estimate) by O and B (the event we know for certain has occurred) by F: P(O|F)=P(OF)P(F)

    Now F={5}, so P(F)=16.

    From part (a) we found that P(FO)=16. Thus,

    P(O|F)=P(OF)P(F)=1616=1

Just as we did not need the computational formula in this example, we do not need it when the information is presented in a two-way contingency table, as in the next example.

Example 2

In a sample of 902 individuals under 40 who were or had previously been married, each person was classified according to gender and age at first marriage. The results are summarized in the following two-way classification table, where the meaning of the labels is:

  1. M: male
  2. F: female
  3. E: a teenager when first married
  4. W: in one’s twenties when first married
  5. H: in one’s thirties when first married
E W H Total
M 43 293 114 450
F 82 299 71 452
Total 125 592 185 902

The numbers in the first row mean that 43 people in the sample were men who were first married in their teens, 293 were men who were first married in their twenties, 114 men who were first married in their thirties, and a total of 450 people in the sample were men. Similarly for the numbers in the second row. The numbers in the last row mean that, irrespective of gender, 125 people in the sample were married in their teens, 592 in their twenties, 185 in their thirties, and that there were 902 people in the sample in all. Suppose that the proportions in the sample accurately reflect those in the population of all individuals in the population who are under 40 and who are or have previously been married. Suppose such a person is selected at random.

  1. Find the probability that the individual selected was a teenager at first marriage.
  2. Find the probability that the individual selected was a teenager at first marriage, given that the person is male.

Solution

It is natural to let E also denote the event that the person selected was a teenager at first marriage and to let M denote the event that the person selected is male.

  1. According to the table the proportion of individuals in the sample who were in their teens at their first marriage is 125902. This is the relative frequency of such people in the population, hence P(E)=1259020.139 or about 14%.
  2. Since it is known that the person selected is male, all the females may be removed from consideration, so that only the row in the table corresponding to men in the sample applies:
    E W H Total
    M 43 293 114 450

    The proportion of males in the sample who were in their teens at their first marriage is 43450. This is the relative frequency of such people in the population of males, hence P(E|M)=434500.096 or about 10%.

Example 3

Suppose that in an adult population the proportion of people who are both overweight and suffer hypertension is 0.09; the proportion of people who are not overweight but suffer hypertension is 0.11; the proportion of people who are overweight but do not suffer hypertension is 0.02; and the proportion of people who are neither overweight nor suffer hypertension is 0.78. An adult is randomly selected from this population.
  1. Find the probability that the person selected suffers hypertension given that they are overweight.
  2. Find the probability that the selected person suffers hypertension given that they are not overweight.
  3. Compare the two probabilities just found to give an answer to the question as to whether overweight people tend to suffer from hypertension.

Solution

Let H denote the event “the person selected suffers hypertension.” Let O denote the event “the person selected is overweight.” The probability information given in the problem may be organized into the following contingency table:

O Oc
H 0.09 0.11
Hc 0.02 0.78
  1. Using the formula in the definition of conditional probability,

    P(H|O)=P(HO)P(O)=0.090.09+0.020.8182

  2. Using the formula in the definition of conditional probability,

    P(H|OC)=P(HOC)P(OC)=0.110.11+0.780.1236

  3. P(H|O)0.8182 is over six times as large as P(H|OC)0.1236, which indicates a much higher rate of hypertension among people who are overweight than among people who are not overweight. It might be interesting to note that a direct comparison of P(HO)=0.09 and P(HOC)=0.11 does not answer the same question.

Try it Now 1

The table below shows the number of survey subjects who have received and not received a speeding ticket in the last year, and the color of their car. Find the probability that a randomly chosen person:

  1. Has a speeding ticket given they have a red car
  2. Has a red car given they have a speeding ticket
Speeding ticket No speeding ticket Total
Red car 15 135 150
Not red car 45 470 515
Total 60 605 665
Hint 1 (click to Show/Hide)

Both of these questions be answered by zooming in on the given information and taking the appropriate ratio. For (a) we would want to look at just the row for having a red car. For (b) we would want to look at just the column of having a speeding ticket.

Answer (click to Show/Hide)
  1. Since we know the person has a red car, we are only considering the 150 people in the first row of the table. Of those, 15 have a speeding ticket, so

    P(T|R)=15150=0.1

  2. Since we know the person has a speeding ticket, we are only considering the 60 people in the first column of the table. Of those, 15 have a red car, so

    P(R|T)=1560=0.25

Notice from the Try It Now 1 that P(R|T) is not equal to P(T|R).

These kinds of conditional probabilities are what insurance companies use to determine your insurance rates. They look at the conditional probability of you having accident, given your age, your car, your car color, your driving history, etc., and price your policy based on that likelihood.

Multiplication Rule

If we rewrite the conditional probability formula we have another tool to answer questions about probability
involving “and“.

General Multiplication Rule

Given that A and B are both events that have probabilities that are not zero we can say the following about the probability of A and B together.

P(AB)=P(B|A)P(A)

P(AB)=P(A|B)P(B)

In the definition above please note this does not mean we are saying PA and PB are equal to each other (or the conditional probabilities listed)

This manipulation of the conditional probability formula gives us another way of finding the probability of A and B (this also means the intersection of A with B). It is not always needed when computing the probability of an intersection (the Addition Rule also can be manipulated or if we know the events A and B are independent we can use the rule from the previous section).

Example 4

If you pull 2 cards out of a deck one after the other and keep both, then what is the probability that both
are spades?


Solution

Let A be the event that the first card drawn is a space and B be the event that the second card drawn is a space. We are looking to find the probabilty that both A and B occur: P(AB).

The probability that the first card is a spade is P(A)=1352.

If the first card drawn is a spade we can find the probability of the second card is a spade by counting the total number of spades left and dividing it by the total number of cards left. The probability that the second card is a spade, given the first was a spade, is P(B|A)=1251, since there is one less spade in the deck, and one less total cards.

The probability that both cards are spades is:

P(AB)=P(B|A)P(A)P(AB)=12511352P(AB)=1562652P(AB)0.0588

Video Solution of the Example (4 mins 28 secs – CC)

Example 5

If you draw two cards from a deck (one after the other), what is the probability that you will get the Ace of Diamonds and a black card?


Solution

You can satisfy this condition by having Case A or Case B, as follows:

  • Case A: you can get the Ace of Diamonds first and then a black card
  • Case B: you can get a black card first and then the Ace of Diamonds

Let A1 be the event we draw the Ace of Diamonds on the first card, A2 be the event we draw the Ace of Diamonds on the second draw, B1 be the event we draw a black card on the first card, and B2 be the event we draw a black card on the second draw.

Let’s calculate the probability of Case A. The probability that the first card is the Ace of Diamonds is P(A1)=152. The probability that the second card is black given that the first card is the Ace of Diamonds is P(B2|A1)=2651 because 26 of the remaining 51 cards are black.

The probability for Case A is:

P(A1B2)=P(B2|A1)P(A1)P(A1B2)=2651152P(A1B2)=1102

Now for Case B: the probability that the first card is black is P(B1)=2652. The probability that the second card is the Ace of Diamonds given that the first card is black is P(A2|B1)=151.

The probability of Case B is therefore

P(B1A2)=P(A2|B1)P(B1)P(B1A2)=13512652P(B1A2)=1102

Recall that the probability of A or B is P(AB)=P(A)+P(B)P(AB). In this problem, P(AB)=0 since the first card cannot be the Ace of Diamonds and be a black card. Therefore, the probability of Case A or Case B is 1102+1102=2102. The probability that you will get the Ace of Diamonds and a black card when drawing two cards from a deck is 2102.

Video Solution of the Example (1 min 55 secs – CC) Another example starts after the 1:55 mark of the video.

Try it Now 2

In your drawer you have 10 pairs of socks, 6 of which are white. If you reach in and randomly grab two pairs of socks, what is the probability that both are white?

Hint 1 (click to Show/Hide)

In order for both to be white we have to have the first drawn to be white and then given that the first is white the second sock drawn is white as well. The key here is that given piece as it tells us we are dealing with conditional probability.

Answer (click to Show/Hide)

Let A be the event we draw a white sock on the first draw and B the event we draw a white sock on the second draw. We are looking to find P(A and B)

Now the multiplication rule can help us as it can be rewritten as: P(A and B)=P(B|A)P(A)

P(A)=610, since we have 6 white socks out of the ten when we draw the first sock from the drawer.

P(B|A)=59, since we are assuming A has occured (we already drawn a sock from the drawer and it was white). The first sock being white means we now have five white socks out of the nine remaining socks in the drawer.

Put the two above together into the formula to get: P(A and B)=P(A)P(B|A)=(610)(59)=13.

Let us put our knowledge of conditional probability, complements, and the multiplication rule to a test in this next Try It Now.

Try it Now 3

The probability that event A occurs is .63. The probability that event B occurs is .45. The probability that both events occur is .10.

Find the following probabilities by using the definition of conditional probability and Venn Diagrams:

  1. P(A|B)
  2. P(B|A)
  3. P(AC|B)
  4. P(B|AC)
  5. P(AC|BC)
Hint 1 (click to Show/Hide)

Start with writing out the given information in probability notation. You will also need the following three formulas:

P(A|B)=P(AB)P(B)P(AB)=P(B|A)P(A)P(AC)=1P(A)

Hint 2 (click to Show/Hide)

If you are stuck try writing out the definition of conditional probabilty for each one to see if it leads to a solution with the given information you already have.

Answer (click to Show/Hide)

The answers along with formulas/steps are given below. To see an alternative approach using Venn Diagrams see the video link much further below.

  1. P(A|B)=P(AB)P(B)=0.100.45=29
  2. P(B|A)=P(AB)P(A)=0.100.63=1063
  3. P(AC|B)=1P(A|B)=129=79: The complement rule was used, but note how the given information doesn’t change. Only the event of
    interest in the conditional probability has the complement taken.
  4. This one takes many different steps to find expressions we are able to use. Started with the general conditional probability formula, but then used the complement rule in the denominator and the multiplication rule in the numerator to eventually get a form we can evaluate from our work in c above.

    P(B|AC)=P(ACB)P(AC)P(B|AC)=P(AC|B)P(B)1P(A)P(B|AC)=79920163100P(B|AC)=72010037P(B|AC)=3537

  5. This one also starts with applying the formula for consitional probability and the use of the multiplication rule and complement rule:

    P(AC|BC)=P(ACBC)P(BC)P(AC|BC)=P(BC|AC)P(AC)1P(B)P(AC|BC)=[1P(B|AC)][1P(A)]1P(B)P(AC|BC)=[13537][163100]1920P(AC|BC)=[237][37100]1120P(AC|BC)=21002011P(AC|BC)=25111P(AC|BC)=255

Now parts d and e are a challenge to construct on your own from just the formulas. There is approach for each of these using a more visual construction for the formulas to get the final result. Watch the Try It Now Video (10 mins 55 secs – CC) for an explanation of how it works.

Independent Events

With conditional probability we saw how some events can often influence each other. However, sometimes events occur without affecting the likelihood of other events. We call these independent events.

Consider an example related to insurance company assessing risks on a home insurance policy:

A: A house experiences storm damage this year
B: The same house had a burglary last year

These events are likely independent because a past burglary doesn’t influence the chances of future storm damage.

Although typically we expect the conditional probability P(A|B) to be different from the probability P(A), it does not have to be true. When P(A|B)=P(A), the occurrence of B has no effect on the likelihood of A. Whether or not the event B has occurred is independent of the event A.

Independent

Events A and B are independent if P(A|B)=P(A).

If A and B are not independent we say they are dependent.

Using algebra it can be shown that the equality P(A|B)=P(A) holds if and only if the equality P(AB)=P(A)P(B) holds, which in turn is true if and only if P(B|A)=P(B).

The formula in the definition has two practical but exactly opposite uses:

  1. In a situation in which we can compute all three probabilities P(A), P(B), and P(AB), it is used to check whether or not the events A and B are independent:
    • If P(AB)=P(A)P(B), then A and B are independent.
    • If P(AB)P(A)P(B), then A and B are not independent.
  2. In a situation in which each of P(A) and P(B) can be computed and it is known that A and B are independent, then we can compute P(AB) by multiplying together P(A) and P(B): P(AB)=P(A)P(B).

Example 6

A single fair die is rolled. Let A={3} and B={1,3,5}. Are A and B independent?


Solution

In this example we can compute all three probabilities P(A)=16, P(B)=12, and P(AB)=16.

The product P(A)P(B)=1612=112 is not the same number as P(AB)=16, the events A and B are not independent.

Example 7

Many diagnostic tests for detecting diseases do not test for the disease directly but for a chemical or biological product of the disease, hence are not perfectly reliable. The sensitivity of a test is the probability that the test will be positive when administered to a person who has the disease. The higher the sensitivity, the greater the detection rate and the lower the false negative rate.

Suppose the sensitivity of a diagnostic procedure to test whether a person has a particular disease is 92%. A person who actually has the disease is tested for it using this procedure by two independent laboratories.

  1. What is the probability that both test results will be positive?
  2. What is the probability that at least one of the two test results will be positive?

Solution

  1. Let A1 denote the event “the test by the first laboratory is positive” and let A2 denote the event “the test by the second laboratory is positive.” Since A1 and A2 are independent,

    P(A1A2)=P(A1)P(A2)=0.920.92=0.8464

  2. Using the Additive Rule for Probability and the probability just computed,

    P(A1A2)=P(A1)+P(A2)P(A1A2)P(A1A2)=0.92+0.920.8464P(A1A2)=0.9936

Example 8

The specificity of a diagnostic test for a disease is the probability that the test will be negative when administered to a person who does not have the disease. The higher the specificity, the lower the false positive rate.

Suppose the specificity of a diagnostic procedure to test whether a person has a particular disease is 89%.

  1. A person who does not have the disease is tested for it using this procedure. What is the probability that the test result will be positive?
  2. A person who does not have the disease is tested for it by two independent laboratories using this procedure. What is the probability that both test results will be positive?

Solution

  1. Let B denote the event “the test result is positive.” The complement of B is that the test result is negative, and has probability the specificity of the test, 0.89. Thus

    P(B)=1P(BC)=10.89=0.11

  2. Let B1 denote the event “the test by the first laboratory is positive” and let B2 denote the event “the test by the second laboratory is positive.” Since B1 and B2 are independent, by part (a) of the example

    P(B1B2)=P(B1)P(B2)=0.110.11=0.0121

Example 9

A jar contains 10 marbles, 7 black and 3 white. Two marbles are drawn without replacement, which means that the first one is not put back before the second one is drawn.

  1. What is the probability that both marbles are black?
  2. What is the probability that exactly one marble is black?
  3. What is the probability that at least one marble is black?

Solution

A tree diagram for the situation of drawing one marble after the other without replacement is shown in “Tree Diagram for Drawing Two Marbles”. The circle and rectangle will be explained later, and should be ignored for now.

Tree Diagram with first branches going to B and W. Branch going to B has 7/10, branch going to W has 3/10. From W branches to B and W with value 7/9 on branch to B and 2/9 on branch to W. From B branches to B and W with 6/9 on B branch and 3/9 on W branch. Shows probabilitys of the two nodes in branches as described in work.
Tree Diagram for Drawing Two Marbles
The numbers on the two leftmost branches are the probabilities of getting either a black marble, 7 out of 10, or a white marble, 3 out of 10, on the first draw. The number on each remaining branch is the probability of the event corresponding to the node on the right end of the branch occurring, given that the event corresponding to the node on the left end of the branch has occurred. Thus for the top branch, connecting the two Bs, it is P(B2|B1), where B1 denotes the event “the first marble drawn is black” and B2 denotes the event “the second marble drawn is black.” Since after drawing a black marble out of the jar there are 9 marbles left, of which 6 are black, this probability is 69.

The number to the right of each final node is computed as shown, using the principle that if the formula in the Conditional Rule for Probability is multiplied by P(B), then the result is

P(BA)=P(B)P(A|B)

  1. The event “both marbles are black” is B1B2 and corresponds to the top right node in the tree, which has been circled. Thus as indicated there, it is:

    P(B1B2)=P(B1)P(B2)=710690.47

  2. The event “exactly one marble is black” corresponds to the two nodes of the tree enclosed by the rectangle. The events that correspond to these two nodes are mutually exclusive: black followed by white B1W2 is incompatible with white followed by black W1B2. Thus in accordance with the Additive Rule for Probability we merely add the two probabilities next to these nodes. Thus the probability of drawing exactly one black marble in two tries is:

    P(B1W2)+P(W1B2)=0.23+0.23=0.46

  3. The event “at least one marble is black” corresponds to the three nodes of the tree enclosed by either the circle or the rectangle. The events that correspond to these nodes are mutually exclusive, so as in part (b) we merely add the probabilities next to these nodes. Thus the probability of drawing at least one black marble in two tries is 0.47+0.23+0.23=0.93.

    P(B1W2)+P(W1B2)+P(B1B2)=0.23+0.23+0.47=0.93

    Of course, this answer could have been found more easily using the Probability Law for Complements, simply subtracting the probability of the complementary event, “two white marbles are drawn,” from 1 to obtain 1−0.07=0.93.

Try it Now 4

Are these events independent?

  1. A fair coin is tossed two times. The two events are (1) first toss is a head and (2) second toss is a head.
  2. The two events (1) “It will rain tomorrow in Tucson” and (2) “It will rain tomorrow in Phoenix” (a city near Tucson).
  3. You draw a red card from a deck, then draw a second card without replacing the first.
  4. A coin is tossed and you record Heads or Tails, and you roll a die and record the outcome.
Answer (click to Show/Hide)
  1. The probability that a head comes up on the second toss is 1/2 regardless of whether or not a head came up on the first toss, so these events are independent.
  2. These events are not independent because it is more likely that it will rain in Phoenix on days it rains in Tucson than on days it does not.
  3. The probability of the second card depends on what the first card drawn is, so these events are not independent.
  4. These are independent as the outcome of the coin toss would not change based on the outcome of the dice roll.

Try it Now 5

A card is pulled from a deck of cards and noted. The card is then replaced, the deck is shuffled, and a second card is removed and noted. What is the probability that both cards are Aces?

Hint 1 (click to Show/Hide)

Can we state the events of drawing those two cards are independent? If we can then we can use the multiplication rule for independent events. If we can’t establish independence, then we have to try something different.

Answer (click to Show/Hide)

Since the second draw is made after replacing the first card, these events are independent. The probability of an ace on each draw is 452=113, so the probability of an Ace on both draws is 113113=1169.


WARNING!!!! Independent ≠ Disjoint (mutually exclusive)

The concepts of independent and mutually exclusive events are often confused.

Mutually Exclusive Events:
  • These are events that cannot happen at the same time.
  • If one occurs, the other cannot occur.

Independent Events:

  • The occurrence of one event doesn’t affect the probability of the other event.
  • Knowing one event happened doesn’t change the likelihood of the other event.

For example let us take a standard 52-card deck. We will define two events

Event B: Drawing a black card
Event H: Drawing a heart

These events are mutually exclusive, since a card can’t be both black and a heart (red) at the same time.

PH=1352=14 , but if we know the card is black, PH|B=0.

This shows that knowing the card is black changes the probability of it being a heart. Therefore, these events are not independent.

Remember:

  • For independent events: PA|B=PA
  • For mutually exclusive events: PAB=0

Exercises


  1. For two events A and B, P(A)=0.73, P(B)=0.48, and P(A∩B)=0.29.
    1. Find P(A|B).
    2. Find P(B|A).
    3. Determine whether or not A and B are independent.
    Answer (click to Show/Hide)

    This problem uses the formula for conditional probability and the definition of independent events.

    The formula for conditional probability of event A given event B is:
    P(A|B)=P(AB)P(B)

    a. Find P(A|B)

    We substitute the given values into the formula.
    P(A|B)=0.290.480.604

    Answer (a): P(A|B)0.604.

    b. Find P(B|A)

    We use the corresponding formula for P(B|A).
    P(B|A)=P(AB)P(A)=0.290.730.397

    Answer (b): P(B|A)0.397.

    c. Determine whether or not A and B are independent.

    Two events A and B are independent if P(A|B)=P(A). We can use our result from part (a) to check if this condition is met.

    Step 1: Compare P(A|B) with P(A).
    From part (a), we found that P(A|B)0.604.
    We are given that P(A)=0.73.
    0.6040.73

    Since P(A|B)P(A), the events are not independent.

    Answer (c): A and B are not independent because P(A|B)P(A).

  2. For two events A and B, P(A)=0.26, P(B)=0.37, and P(A∩B)=0.11.
    1. Find P(A|B).
    2. Find P(B|A).
    3. Determine whether or not A and B are independent.
    Answer (click to Show/Hide)

    This problem uses the formula for conditional probability and the definition of independent events.

    The formula for conditional probability of event A given event B is:
    P(A|B)=P(AB)P(B)

    a. Find P(A|B)

    We substitute the given values into the formula.
    P(A|B)=0.110.370.297

    Answer (a): P(A|B)0.297.

    b. Find P(B|A)

    We use the corresponding formula for P(B|A).
    P(B|A)=P(AB)P(A)=0.110.260.423

    Answer (b): P(B|A)0.423.

    c. Determine whether or not A and B are independent.

    Two events A and B are independent if P(A|B)=P(A). We can use our result from part (a) to check if this condition is met.

    Step 1: Compare P(A|B) with P(A).
    From part (a), we found that P(A|B)0.297.
    We are given that P(A)=0.26.
    0.2970.26

    Since P(A|B)P(A), the events are not independent.

    Answer (c): A and B are not independent because P(A|B)P(A).

  3. For two events A and B, P(A)=0.25, P(B)=0.6, and P(A∩B)=0.15.
    1. Find P(A|B).
    2. Find P(B|A).
    3. Determine whether or not A and B are independent.
    4. Determine whether or not A and B are mutually exclusive.
    Answer (click to Show/Hide)

    This problem uses the formulas for conditional probability, independent events, and mutually exclusive events.

    a. Find P(A|B)

    The formula for conditional probability is P(A|B)=P(AB)P(B).
    We substitute the given values into the formula.
    P(A|B)=0.150.6=0.25

    Answer (a): P(A|B)=0.25.

    b. Find P(B|A)

    We use the corresponding formula for P(B|A).
    P(B|A)=P(AB)P(A)=0.150.25=0.6

    Answer (b): P(B|A)=0.6.

    c. Determine whether or not A and B are independent.

    Two events A and B are independent if P(A|B)=P(A).
    From part (a), we found that P(A|B)=0.25.
    We are given that P(A)=0.25.
    Since 0.25=0.25, the events are independent.

    Answer (c): A and B are independent because P(A|B)=P(A).

    d. Determine whether or not A and B are mutually exclusive.

    Two events are mutually exclusive if they cannot happen at the same time, which means their intersection is zero: P(AB)=0.
    We are given that P(AB)=0.15.
    Since 0.150, the events are not mutually exclusive.

    Answer (d): A and B are not mutually exclusive because their probability of occurring together is greater than 0.

  4. Compute the following probabilities in connection with the roll of a single fair die.
    1. The probability that the roll is even.
    2. The probability that the roll is even, given that it is not a two.
    3. The probability that the roll is even, given that it is not a one.
    Answer (click to Show/Hide)

    The sample space for a single fair die roll is S = {1, 2, 3, 4, 5, 6}.

    a. The probability that the roll is even.

    This is a basic probability problem. The event E is “the roll is even.”

    Step 1: Find the number of favorable outcomes.
    The even numbers in the sample space are {2, 4, 6}. There are 3 favorable outcomes.

    Step 2: Calculate the probability.
    The total number of outcomes is 6.
    P(Even)=Number of even outcomesTotal number of outcomes=36=12

    Answer (a): The probability of rolling an even number is 1/2.

    b. The probability that the roll is even, given that it is not a two.

    Let A be the event “the roll is even” and B be the event “the roll is not a two.” We want to find P(A|B).

    The formal definition for conditional probability is: P(A|B)=P(AB)P(B)

    Step 1: Find P(B), the probability the roll is not a two.
    The outcomes for “not a two” are {1, 3, 4, 5, 6}. There are 5 such outcomes.
    P(B)=P(Not a 2)=56

    Step 2: Find P(A∩B), the probability the roll is even AND not a two.
    The outcomes that are both even and not a two are {4, 6}. There are 2 such outcomes.
    P(AB)=P(Even and Not a 2)=26

    Step 3: Calculate P(A|B).
    P(A|B)=P(AB)P(B)=2/65/6=25

    Alternative Method (Reduced Sample Space):
    The given condition, “the roll is not a two,” reduces our sample space to S’ = {1, 3, 4, 5, 6}. Within this new space of 5 outcomes, the favorable (even) outcomes are {4, 6}. There are 2 favorable outcomes. Therefore, the probability is 2/5.

    Answer (b): The probability is 2/5.

    c. The probability that the roll is even, given that it is not a one.

    Let A be the event “the roll is even” and C be the event “the roll is not a one.” We want to find P(A|C).

    Step 1: Find P(C), the probability the roll is not a one.
    The outcomes for “not a one” are {2, 3, 4, 5, 6}. There are 5 such outcomes.
    P(C)=P(Not a 1)=56

    Step 2: Find P(A∩C), the probability the roll is even AND not a one.
    The outcomes that are both even and not a one are {2, 4, 6}. There are 3 such outcomes.
    P(AC)=P(Even and Not a 1)=36

    Step 3: Calculate P(A|C).
    P(A|C)=P(AC)P(C)=3/65/6=35

    Alternative Method (Reduced Sample Space):
    The given condition, “the roll is not a one,” reduces our sample space to S” = {2, 3, 4, 5, 6}. Within this new space of 5 outcomes, the favorable (even) outcomes are {2, 4, 6}. There are 3 favorable outcomes. Therefore, the probability is 3/5.

    Answer (c): The probability is 3/5.

  5. Compute the following probabilities in connection with two tosses of a fair coin.
    1. The probability that the second toss is heads.
    2. The probability that the second toss is heads, given that the first toss is heads.
    3. The probability that the second toss is heads, given that at least one of the two tosses is heads.
    Answer (click to Show/Hide)

    The sample space for two tosses of a fair coin is S = {HH, HT, TH, TT}, where H denotes heads and T denotes tails. There are 4 total equally likely outcomes.

    a. The probability that the second toss is heads.

    This is a basic probability problem. Let A be the event “the second toss is heads.”

    Step 1: Find the number of favorable outcomes.
    The outcomes where the second toss is heads are {HH, TH}. There are 2 favorable outcomes.

    Step 2: Calculate the probability.
    The total number of outcomes is 4.
    P(A)=Number of outcomes with second toss HTotal number of outcomes=24=12

    Answer (a): The probability that the second toss is heads is 1/2.

    b. The probability that the second toss is heads, given that the first toss is heads.

    Let A be the event “the second toss is heads” and B be the event “the first toss is heads.” We want to find P(A|B).

    The formal definition for conditional probability is: P(A|B)=P(AB)P(B)

    Step 1: Find P(B), the probability the first toss is heads.
    The outcomes for “first toss is heads” are {HH, HT}. There are 2 such outcomes.
    P(B)=24=12

    Step 2: Find P(A∩B), the probability the first toss is heads AND the second toss is heads.
    The only outcome that satisfies this is {HH}. There is 1 such outcome.
    P(AB)=14

    Step 3: Calculate P(A|B).
    P(A|B)=P(AB)P(B)=1/42/4=12

    Alternative Method (Reduced Sample Space):
    The given condition, “the first toss is heads,” reduces our sample space to S’ = {HH, HT}. Within this new space of 2 outcomes, the favorable outcome (second toss is heads) is {HH}. There is 1 favorable outcome. Therefore, the probability is 1/2.

    Answer (b): The probability is 1/2.

    c. The probability that the second toss is heads, given that at least one of the two tosses is heads.

    Let A be the event “the second toss is heads” and C be the event “at least one toss is heads.” We want to find P(A|C).

    Step 1: Find P(C), the probability at least one toss is heads.
    The outcomes for “at least one heads” are {HH, HT, TH}. There are 3 such outcomes.
    P(C)=34

    Step 2: Find P(A∩C), the probability the second toss is heads AND at least one toss is heads.
    The outcomes that satisfy both conditions are {HH, TH}. There are 2 such outcomes.
    P(AC)=24

    Step 3: Calculate P(A|C).
    P(A|C)=P(AC)P(C)=2/43/4=23

    Alternative Method (Reduced Sample Space):
    The given condition, “at least one of the two tosses is heads,” reduces our sample space to S” = {HH, HT, TH}. Within this new space of 3 outcomes, the favorable outcomes (second toss is heads) are {HH, TH}. There are 2 favorable outcomes. Therefore, the probability is 2/3.

    Answer (c): The probability is 2/3.

  6. A special deck of 16 cards has 4 that are blue, 4 yellow, 4 green, and 4 red. The four cards of each color are numbered from one to four. A single card is drawn at random. Find the following probabilities.
    1. The probability that the card drawn is red.
    2. The probability that the card is red, given that it is not green.
    3. The probability that the card is red, given that it is neither red nor yellow.
    4. The probability that the card is red, given that it is not a four.
    Answer (click to Show/Hide)

    The total sample space has 16 cards.

    a. The probability that the card drawn is red.
    There are 4 red cards out of 16 total cards.
    P(Red)=416=14
    Answer (a): The probability is 1/4.

    b. The probability that the card is red, given that it is not green.
    The condition “not green” reduces the sample space to 12 cards (4 red, 4 blue, 4 yellow). Within this smaller group, there are 4 red cards.
    P(Red|Not Green)=412=13
    Answer (b): The probability is 1/3.

    c. The probability that the card is red, given that it is neither red nor yellow.
    The condition “neither red nor yellow” reduces the sample space to the 8 cards that are blue or green. Within this group, there are 0 red cards.
    P(Red|Neither Red nor Yellow)=08=0
    Answer (c): The probability is 0.

    d. The probability that the card is red, given that it is not a four.
    The condition “not a four” reduces the sample space to 12 cards (the cards numbered 1, 2, and 3 for each of the four colors). Within this group, there are 3 red cards (the red 1, red 2, and red 3).
    P(Red|Not a Four)=312=14
    Answer (d): The probability is 1/4.

  7. A special deck of 16 cards has 4 that are blue, 4 yellow, 4 green, and 4 red. The four cards of each color are numbered from one to four. A single card is drawn at random. Find the following probabilities.
    1. The probability that the card drawn is a two or a four.
    2. The probability that the card is a two or a four, given that it is not a one.
    3. The probability that the card is a two or a four, given that it is either a two or a three.
    4. The probability that the card is a two or a four, given that it is red or green.
    Answer (click to Show/Hide)

    The total sample space has 16 cards.

    a. The probability that the card drawn is a two or a four.
    There are 4 cards numbered “two” and 4 cards numbered “four,” for a total of 8 favorable outcomes out of 16 total cards.
    P(Two or Four)=816=12
    Answer (a): The probability is 1/2.

    b. The probability that the card is a two or a four, given that it is not a one.
    The condition “not a one” reduces the sample space to 12 cards (all cards except the four “ones”). Within this group, there are still 4 “twos” and 4 “fours,” making 8 favorable outcomes.
    P(Two or Four|Not a One)=812=23
    Answer (b): The probability is 2/3.

    c. The probability that the card is a two or a four, given that it is either a two or a three.
    The condition “either a two or a three” reduces the sample space to 8 cards (the four “twos” and the four “threes”). Within this group, only the 4 “twos” are favorable outcomes.
    P(Two or Four|Two or Three)=48=12
    Answer (c): The probability is 1/2.

    d. The probability that the card is a two or a four, given that it is red or green.
    The condition “red or green” reduces the sample space to 8 cards (the 4 red and 4 green cards). Within this group, there are 2 favorable outcomes from the red cards (the red 2 and 4) and 2 from the green cards (the green 2 and 4), for a total of 4 favorable outcomes.
    P(Two or Four|Red or Green)=48=12
    Answer (d): The probability is 1/2.

  8. A random experiment gave rise to the two-way contingency table shown. Use it to compute the probabilities indicated.
    R S
    A 0.12 0.18
    B 0.28 0.42
    1. P(A), P(R), P(A∩R).
    2. Based on the answer to (a), determine whether or not the events A and R are independent.
    3. Based on the answer to (b), determine whether or not P(A|R) can be predicted without any computation. If so, make the prediction. In any case, compute P(A|R) using the Rule for Conditional Probability.
    Answer (click to Show/Hide)
    First, it’s helpful to add totals to the contingency table.
    R S Row Total
    A 0.12 0.18 0.30
    B 0.28 0.42 0.70
    Column Total 0.40 0.60 1.00

    a. Find P(A), P(R), P(A∩R).

    P(A) is the total probability for row A, which is the sum of the probabilities in that row.
    P(A)=0.12+0.18=0.30

    P(R) is the total probability for column R, which is the sum of the probabilities in that column.
    P(R)=0.12+0.28=0.40

    P(AR) is the probability of the intersection of A and R, which is the value in the cell where they meet.
    P(AR)=0.12

    Answer (a): P(A)=0.30, P(R)=0.40, and P(AR)=0.12.

    b. Based on the answer to (a), determine whether or not the events A and R are independent.

    Two events are independent if P(AR)=P(A)×P(R). Let’s check this condition.
    P(A)×P(R)=0.30×0.40=0.12
    We found that P(AR) is also 0.12. Since the values are equal, the events are independent.

    Answer (b): The events A and R are independent.

    c. Based on the answer to (b), determine whether or not P(A|R) can be predicted without any computation.

    Yes, it can be predicted. Because we found that A and R are independent events, we know that the occurrence of R does not affect the probability of A. Therefore, P(A|R) must be equal to P(A). The prediction is that P(A|R)=0.30.

    Now we compute it using the rule for conditional probability to confirm:
    P(A|R)=P(AR)P(R)=0.120.40=0.30
    The computation matches the prediction.

    Answer (c): Yes, it can be predicted to be 0.30. The computation confirms this result.

  9. A random experiment gave rise to the two-way contingency table shown. Use it to compute the probabilities indicated.
    R S
    A 0.13 0.07
    B 0.61 0.19
    1. P(A), P(R), P(A∩R).
    2. Based on the answer to (a), determine whether or not the events A and R are independent.
    3. Based on the answer to (b), determine whether or not P(A|R) can be predicted without any computation. If so, make the prediction. In any case, compute P(A|R) using the Rule for Conditional Probability.
    Answer (click to Show/Hide)
    First, it’s helpful to add totals to the contingency table.
    R S Row Total
    A 0.13 0.07 0.20
    B 0.61 0.19 0.80
    Column Total 0.74 0.26 1.00

    a. Find P(A), P(R), P(A∩R).

    P(A) is the total probability for row A.
    P(A)=0.13+0.07=0.20

    P(R) is the total probability for column R.
    P(R)=0.13+0.61=0.74

    P(AR) is the value in the cell where A and R intersect.
    P(AR)=0.13

    Answer (a): P(A)=0.20, P(R)=0.74, and P(AR)=0.13.

    b. Based on the answer to (a), determine whether or not the events A and R are independent.

    Two events are independent if P(AR)=P(A)×P(R). Let’s check this condition.
    P(A)×P(R)=0.20×0.74=0.148
    We found that P(AR) is 0.13. Since 0.130.148, the events are not independent.

    Answer (b): The events A and R are not independent.

    c. Based on the answer to (b), determine whether or not P(A|R) can be predicted without any computation.

    No, it cannot be predicted. Because the events A and R are not independent, the probability of A is affected by the occurrence of R. Therefore, P(A|R) will not be equal to P(A).

    We must compute it using the rule for conditional probability:
    P(A|R)=P(AR)P(R)=0.130.740.176

    Answer (c): No, it cannot be predicted. The computed value is approximately 0.176.

  10. Suppose for events A and B in a random experiment P(A)=0.70 and P(B)=0.30. Compute the indicated probability, or explain why there is not enough information to do so.
    1. P(A∩B).
    2. P(A∩B), with the extra information that A and B are independent.
    3. P(A∩B), with the extra information that A and B are mutually exclusive.
    Answer (click to Show/Hide)

    a. P(A∩B).

    With only the probabilities of the individual events, P(A) and P(B), it is impossible to compute the probability of their intersection, P(AB). We do not know if the events are independent, mutually exclusive, or have some other relationship.

    Answer (a): There is not enough information to compute the probability.

    b. P(A∩B), with the extra information that A and B are independent.

    If events A and B are independent, then the probability of their intersection is the product of their individual probabilities.
    P(AB)=P(A)×P(B)

    Step 1: Substitute the given values.
    P(AB)=0.70×0.30=0.21

    Answer (b): If A and B are independent, P(AB)=0.21.

    c. P(A∩B), with the extra information that A and B are mutually exclusive.

    If two events are mutually exclusive, they cannot both occur at the same time. This means their intersection is empty, and the probability of their intersection is zero.
    P(AB)=0

    Answer (c): If A and B are mutually exclusive, P(AB)=0.

  11. Suppose for events A and B connected to some random experiment, P(A)=0.50 and P(B)=0.50. Compute the indicated probability, or explain why there is not enough information to do so.
    1. P(A∩B).
    2. P(A∩B), with the extra information that A and B are independent.
    3. P(A∩B), with the extra information that A and B are mutually exclusive.
    Answer (click to Show/Hide)

    a. P(A∩B).

    With only the probabilities of the individual events, P(A) and P(B), it is impossible to compute the probability of their intersection, P(AB). We do not know if the events are independent, mutually exclusive, or have some other relationship.

    Answer (a): There is not enough information to compute the probability.

    b. P(A∩B), with the extra information that A and B are independent.

    If events A and B are independent, then the probability of their intersection is the product of their individual probabilities.
    P(AB)=P(A)×P(B)

    Step 1: Substitute the given values.
    P(AB)=0.50×0.50=0.25

    Answer (b): If A and B are independent, P(AB)=0.25.

    c. P(A∩B), with the extra information that A and B are mutually exclusive.

    If two events are mutually exclusive, they cannot both occur at the same time. This means their intersection is empty, and the probability of their intersection is zero.
    P(AB)=0

    Answer (c): If A and B are mutually exclusive, P(AB)=0.

  12. The following two-way contingency table gives the breakdown of the population of patrons at a grocery store according to the number of items purchased and whether or not the patron made an impulse purchase at the checkout counter:
    Number of Items Impulse Purchase
    Made Not Made
    Few 0.01 0.19
    Many 0.04 0.76

    A patron is selected at random. Find each of the following probabilities.

    1. The patron made an impulse purchase.
    2. The patron made an impulse purchase, given that the total number of items purchased was many.
    3. Determine whether or not the events “few purchases” and “made an impulse purchase at the checkout counter” are independent.
    Answer (click to Show/Hide)
    First, we create a new table with row and column totals, which will help in our calculations. Let F be the event “few items,” M be the event “many items,” I be the event “made impulse purchase,” and N be the event “no impulse purchase.”
    I (Made) N (Not Made) Row Total
    F (Few) 0.01 0.19 0.20
    M (Many) 0.04 0.76 0.80
    Column Total 0.05 0.95 1.00

    a. The patron made an impulse purchase.

    This corresponds to the total probability of the “Made” column.
    P(I)=0.01+0.04=0.05

    Answer (a): The probability that a patron made an impulse purchase is 0.05.

    b. The patron made an impulse purchase, given that the total number of items purchased was many.

    This is the conditional probability P(I|M). We use the formula:
    P(I|M)=P(IM)P(M)
    From the table, P(IM)=0.04 and P(M)=0.80.
    P(I|M)=0.040.80=0.05

    Answer (b): The probability is 0.05.

    c. Determine whether or not the events “few purchases” and “made an impulse purchase at the checkout counter” are independent.

    The events F (“few purchases”) and I (“made an impulse purchase”) are independent if P(FI)=P(F)×P(I).

    Step 1: Find the probabilities from the table.

    P(FI)=0.01

    P(F)=0.20

    P(I)=0.05

    Step 2: Check if the condition for independence is met.
    P(F)×P(I)=0.20×0.05=0.01
    Since P(FI) is also 0.01, the condition is met.

    Answer (c): The events are independent.

  13. The sensitivity of a drug test, also known as its true positive rate, is the probability that the test correctly returns a positive result for a person who has actually taken the drug. Suppose that there are two independent tests to detect the presence of a certain type of banned drugs in athletes. One has sensitivity 0.75; the other has sensitivity 0.85. If both are applied to an athlete who has taken this type of drug, what is the chance that their usage will go undetected?
    Answer (click to Show/Hide)

    Let T1 be the event that the first test is positive, and T2 be the event that the second test is positive. The sensitivities are given as probabilities:
    P(T1)=0.75
    P(T2)=0.85

    The drug usage will go undetected if both tests are negative. Let T1C and T2C be the events that the first and second tests are negative, respectively. These are the complements of T1 and T2.

    Step 1: Find the probability of each test being negative.
    The probability of a complement event is 1 minus the probability of the event.
    P(T1C)=1P(T1)=10.75=0.25
    P(T2C)=1P(T2)=10.85=0.15

    Step 2: Find the probability of both tests being negative.
    Since the two tests are independent, the events T1C and T2C are also independent. The probability that both events occur is the product of their individual probabilities.
    P(T1CT2C)=P(T1C)×P(T2C)=0.25×0.15=0.0375

    Answer: The chance that the drug usage will go undetected is 0.0375, or 3.75%.

  14. A person has two lights in their well house to keep the pipes from freezing in winter. They check the lights daily. Each light has probability 0.002 of burning out before it is checked the next day (independently of the other light).
    1. If the lights are wired in parallel one will continue to shine even if the other burns out. In this situation, compute the probability that at least one light will continue to shine for the full 24 hours. Note the greatly increased reliability of the system of two bulbs over that of a single bulb.
    2. If the lights are wired in series neither one will continue to shine even if only one of them burns out. In this situation, compute the probability that at least one light will continue to shine for the full 24 hours. Note the slightly decreased reliability of the system of two bulbs over that of a single bulb.
    Answer (click to Show/Hide)

    Let B1 be the event that the first light burns out, and B2 be the event that the second light burns out. We are given the probabilities:
    P(B1)=0.002
    P(B2)=0.002

    a. Lights Wired in Parallel

    The event “at least one light will continue to shine” is the complement of the event “both lights burn out.” Let’s find the probability that both lights burn out and subtract it from 1.

    Step 1: Find the probability of both lights burning out.
    Since the events are independent, the probability that both B1 and B2 occur is the product of their individual probabilities.
    P(B1B2)=P(B1)×P(B2)=0.002×0.002=0.000004

    Step 2: Use the complement rule.
    The probability of the complement event, “at least one light shines,” is 1 minus the probability that both burn out.
    P(at least one shines)=1P(B1B2)=10.000004=0.999996

    Answer (a): The probability that at least one light will continue to shine is 0.999996, or 99.9996%.

    b. Lights Wired in Series

    If the lights are in series, the system only works if both lights continue to shine. Let B1C and B2C be the events that the first and second lights continue to shine, respectively.

    Step 1: Find the probability of each light continuing to shine.
    This is the complement of the probability that a light burns out.
    P(B1C)=1P(B1)=10.002=0.998
    P(B2C)=1P(B2)=10.002=0.998

    Step 2: Find the probability of both lights continuing to shine.
    Since the events are independent, the probability that both B1C and B2C occur is the product of their individual probabilities.
    P(B1CB2C)=P(B1C)×P(B2C)=0.998×0.998=0.996004

    Answer (b): The probability that at least one light will continue to shine (which means both must shine) is 0.996004, or 99.6004%.

  15. Events A and B are mutually exclusive. Find P(A|B).
    Answer (click to Show/Hide)

    This problem requires understanding the definitions of mutually exclusive events and conditional probability.

    Step 1: Understand Mutually Exclusive Events.
    If two events, A and B, are mutually exclusive, it means that they cannot both happen at the same time. The occurrence of one event precludes the occurrence of the other. Mathematically, this means the probability of their intersection is zero.
    P(AB)=0

    Step 2: Use the Conditional Probability Formula.
    The formula for the conditional probability of A given B is:
    P(A|B)=P(AB)P(B)

    Step 3: Substitute and Solve.
    We substitute the value from Step 1 into the formula from Step 2.
    P(A|B)=0P(B)=0

    This makes intuitive sense: if we know that event B has occurred, then it is impossible for event A to have occurred, since they are mutually exclusive.

    Answer: The probability P(A|B) is 0.

  16. A basketball player makes 60% of the free throws that he attempts, except that if he has just tried and missed a free throw then his chances of making a second one go down to only 30%. Suppose he has just been awarded two free throws.
    1. Find the probability that he makes both.
    2. Find the probability that he makes at least one.
    Answer (click to Show/Hide)

    Let M1 be the event that the player makes the first free throw, and M2 be the event that he makes the second. Let M1C and M2C be the corresponding events that he misses the shots.

    From the problem statement, we can define the following probabilities:

    The probability of making the first shot is P(M1)=0.60.

    The probability of missing the first shot is P(M1C)=10.60=0.40.

    The probability of making the second shot, given he missed the first, is P(M2|M1C)=0.30.

    Since the probability only changes after a miss, the probability of making the second shot, given he made the first, remains unchanged: P(M2|M1)=0.60.

    a. Find the probability that he makes both.

    This is the probability that he makes the first shot AND makes the second shot, P(M1M2). We use the multiplication rule for conditional probability.
    P(M1M2)=P(M1)×P(M2|M1)
    P(M1M2)=0.60×0.60=0.36

    Answer (a): The probability that he makes both free throws is 0.36.

    b. Find the probability that he makes at least one.

    The event “he makes at least one” is the complement of the event “he misses both.” It is easier to calculate the probability that he misses both shots and subtract this from 1.

    Step 1: Find the probability that he misses both shots.
    This is the probability he misses the first shot AND misses the second shot, P(M1CM2C).
    First, we need the probability of missing the second shot, given he missed the first:
    P(M2C|M1C)=1P(M2|M1C)=10.30=0.70
    Now we can find the probability of missing both:
    P(M1CM2C)=P(M1C)×P(M2C|M1C)=0.40×0.70=0.28

    Step 2: Use the complement rule.
    The probability of making at least one shot is 1 minus the probability of missing both.
    P(at least one made)=1P(M1CM2C)=10.28=0.72

    Answer (b): The probability that he makes at least one free throw is 0.72.

  17. Suppose a jar contains 17 red marbles and 32 blue marbles. If you reach in the jar and pull out 2 marbles at random one at a time without replacement, find the probability that both are red.
    Answer (click to Show/Hide)

    This is a conditional probability problem because the outcome of the second draw is dependent on the outcome of the first draw (since the marble is not replaced).

    Let R1 be the event that the first marble drawn is red, and R2 be the event that the second marble drawn is red. We want to find the probability that the first is red AND the second is red, which can be written as P(R1R2).

    We use the multiplication rule for conditional probability:
    P(R1R2)=P(R1)×P(R2|R1)

    Step 1: Find the probability that the first marble is red.
    Initially, there are 17 red marbles and a total of 17 + 32 = 49 marbles.
    P(R1)=1749

    Step 2: Find the probability that the second marble is red, given the first was red.
    After drawing one red marble, there are 16 red marbles remaining and a total of 48 marbles left in the jar.
    P(R2|R1)=1648=13

    Step 3: Calculate the probability of both events occurring.
    Now we multiply the probabilities from the first two steps.
    P(R1R2)=1749×1648=1749×13=17147
    As a decimal, this is approximately 0.1156.

    Answer: The probability that both marbles are red is 17/147, or about 11.6%.

  18. Suppose you write each letter of the alphabet on a different slip of paper and put the slips into a hat. If you pull out two slips at random (one at a time without replacement), find the probability that both are vowels (a, e, i, o , or u).
    Answer (click to Show/Hide)

    This is a conditional probability problem because the outcome of the second draw is dependent on the outcome of the first draw (since the slip is not replaced).

    Let V1 be the event that the first slip drawn is a vowel, and V2 be the event that the second slip drawn is a vowel. We want to find the probability that the first is a vowel AND the second is a vowel, which is P(V1V2).

    We use the multiplication rule for conditional probability:
    P(V1V2)=P(V1)×P(V2|V1)

    Step 1: Find the probability that the first slip is a vowel.
    There are 26 letters in total and 5 of them are vowels.
    P(V1)=526

    Step 2: Find the probability that the second slip is a vowel, given the first was a vowel.
    After drawing one vowel, there are 4 vowels remaining and a total of 25 slips left in the hat.
    P(V2|V1)=425

    Step 3: Calculate the probability of both events occurring.
    Now we multiply the probabilities from the first two steps.
    P(V1V2)=526×425=20650=265
    As a decimal, this is approximately 0.0308.

    Answer: The probability that both slips are vowels is 2/65, or about 3.1%.

  19. A math class consists of 25 students, 14 female and 11 male.  Two students are selected at random to participate in a probability experiment.  Compute the probability that
    1. a male is selected, then a female.
    2. a female is selected, then a male.
    3. two males are selected.
    4. two females are selected.
    5. no males are selected.
    Answer (click to Show/Hide)

    This problem involves conditional probability as the students are selected without replacement. The total number of students is 25.

    a. A male is selected, then a female.

    Let M1 be the event “the first student is male” and F2 be the event “the second student is female.”

    The probability of selecting a male first is P(M1)=1125.

    After a male is chosen, there are 24 students left, 14 of whom are female. The probability of selecting a female second is P(F2|M1)=1424.
    P(M1F2)=1125×1424=154600=77300
    Answer (a): The probability is 77/300.

    b. A female is selected, then a male.

    Let F1 be the event “the first student is female” and M2 be the event “the second student is male.”

    The probability of selecting a female first is P(F1)=1425.

    After a female is chosen, there are 24 students left, 11 of whom are male. The probability of selecting a male second is P(M2|F1)=1124.
    P(F1M2)=1425×1124=154600=77300
    Answer (b): The probability is 77/300.

    c. Two males are selected.

    The probability of selecting a male first is P(M1)=1125.

    After a male is chosen, there are 10 males left out of 24 students. The probability of selecting another male is P(M2|M1)=1024.
    P(M1M2)=1125×1024=110600=1160
    Answer (c): The probability is 11/60.

    d. Two females are selected.

    The probability of selecting a female first is P(F1)=1425.

    After a female is chosen, there are 13 females left out of 24 students. The probability of selecting another female is P(F2|F1)=1324.
    P(F1F2)=1425×1324=182600=91300
    Answer (d): The probability is 91/300.

    e. No males are selected.
    This event is identical to the event “two females are selected.”
    Answer (e): The probability is 91/300.

  20. A math class consists of 25 students, 14 female and 11 male.  Three students are selected at random to participate in a probability experiment.  Compute the probability that
    1. a male is selected, then two females.
    2. a female is selected, then two males.
    3. two females are selected, then one male.
    4. three males are selected.
    5. three females are selected.
    Answer (click to Show/Hide)

    This problem involves conditional probability as the students are selected without replacement from a total of 25 students.

    a. A male is selected, then two females (M, F, F).
    P(M, F, F)=1125×1424×1323=200213800=10016900
    Answer (a): The probability is 1001/6900.

    b. A female is selected, then two males (F, M, M).
    P(F, M, M)=1425×1124×1023=154013800=77690
    Answer (b): The probability is 77/690.

    c. Two females are selected, then one male (F, F, M).
    P(F, F, M)=1425×1324×1123=200213800=10016900
    Answer (c): The probability is 1001/6900.

    d. Three males are selected (M, M, M).
    P(M, M, M)=1125×1024×923=99013800=33460
    Answer (d): The probability is 33/460.

    e. Three females are selected (F, F, F).
    P(F, F, F)=1425×1324×1223=218413800=91575
    Answer (e): The probability is 91/575.

  21. A jar contains 6 red marbles numbered 1 to 6 and 8 blue marbles numbered 1 to 8. A marble is drawn at random from the jar. Find the probability the marble is red or odd-numbered.
    Answer (click to Show/Hide)

    This problem involves the union of two events. Let R be the event that the marble drawn is red, and let O be the event that the marble is odd-numbered. We want to find P(RO).

    The formula for the union of two events is:
    P(RO)=P(R)+P(O)P(RO)

    Step 1: Find the total number of outcomes.
    There are 6 red + 8 blue = 14 marbles in the jar.

    Step 2: Find the probability of drawing a red marble, P(R).
    There are 6 red marbles.
    P(R)=614

    Step 3: Find the probability of drawing an odd-numbered marble, P(O).

    Odd-numbered red marbles: {1, 3, 5} (3 marbles)

    Odd-numbered blue marbles: {1, 3, 5, 7} (4 marbles)
    There are a total of 3 + 4 = 7 odd-numbered marbles.
    P(O)=714

    Step 4: Find the probability of drawing a marble that is both red and odd-numbered, P(R∩O).
    The red marbles with odd numbers are {1, 3, 5}. There are 3 such marbles.
    P(RO)=314

    Step 5: Calculate the final probability.
    Now we use the union formula.
    P(RO)=614+714314=1014=57

    Answer: The probability of drawing a marble that is red or odd-numbered is 5/7.

  22. A jar contains 4 red marbles numbered 1 to 4 and 10 blue marbles numbered 1 to 10. A marble is drawn at random from the jar. Find the probability the marble is blue or even-numbered.
    Answer (click to Show/Hide)

    This problem involves the union of two events. Let B be the event that the marble drawn is blue, and let E be the event that the marble is even-numbered. We want to find P(BE).

    The formula for the union of two events is:
    P(BE)=P(B)+P(E)P(BE)

    Step 1: Find the total number of outcomes.
    There are 4 red + 10 blue = 14 marbles in the jar.

    Step 2: Find the probability of drawing a blue marble, P(B).
    There are 10 blue marbles.
    P(B)=1014

    Step 3: Find the probability of drawing an even-numbered marble, P(E).

    Even-numbered red marbles: {2, 4} (2 marbles)

    Even-numbered blue marbles: {2, 4, 6, 8, 10} (5 marbles)
    There are a total of 2 + 5 = 7 even-numbered marbles.
    P(E)=714

    Step 4: Find the probability of drawing a marble that is both blue and even-numbered, P(B∩E).
    The blue marbles with even numbers are {2, 4, 6, 8, 10}. There are 5 such marbles.
    P(BE)=514

    Step 5: Calculate the final probability.
    Now we use the union formula.
    P(BE)=1014+714514=1214=67

    Answer: The probability of drawing a marble that is blue or even-numbered is 6/7.

  23. A jar contains 5 red marbles numbered 1 to 5 and 8 blue marbles numbered 1 to 8. A marble is drawn at random from the jar. Find the probability the marble is
    1. Even-numbered given that the marble is red.
    2. Red given that the marble is even-numbered.
    Answer (click to Show/Hide)

    This problem involves conditional probability, where we find the probability of an event given that another event has already occurred. The total number of marbles in the jar is 5 + 8 = 13.

    a. Even-numbered given that the marble is red.

    Let E be the event “the marble is even-numbered” and R be the event “the marble is red.” We want to find P(E|R).

    We can solve this by considering the reduced sample space. The condition “the marble is red” means we only need to consider the 5 red marbles.

    Step 1: Identify the new sample space.
    The sample space is the set of red marbles: {R1, R2, R3, R4, R5}. The total number of outcomes in this new space is 5.

    Step 2: Find the number of favorable outcomes within the new sample space.
    The favorable outcomes are the even-numbered marbles within the red marbles, which are {R2, R4}. There are 2 favorable outcomes.

    Step 3: Calculate the probability.
    P(E|R)=Number of even red marblesTotal number of red marbles=25

    Answer (a): The probability is 2/5.

    b. Red given that the marble is even-numbered.

    The condition “the marble is even-numbered” reduces our sample space.

    Step 1: Identify the new sample space.
    The even-numbered marbles are:

    Red: {R2, R4} (2 marbles)

    Blue: {B2, B4, B6, B8} (4 marbles)
    The new sample space is all even-numbered marbles, so the total number of outcomes is 2 + 4 = 6.

    Step 2: Find the number of favorable outcomes within the new sample space.
    The favorable outcomes are the red marbles within this group of even-numbered marbles, which are {R2, R4}. There are 2 favorable outcomes.

    Step 3: Calculate the probability.
    P(R|E)=Number of red even marblesTotal number of even marbles=26=13

    Answer (b): The probability is 1/3.

  24. A jar contains 4 red marbles numbered 1 to 4 and 8 blue marbles numbered 1 to 8. A marble is drawn at random from the jar. Find the probability the marble is
    1. Odd-numbered given that the marble is blue.
    2. Blue given that the marble is odd-numbered.
    Answer (click to Show/Hide)

    This problem involves conditional probability. The total number of marbles in the jar is 4 + 8 = 12.

    a. Odd-numbered given that the marble is blue.

    Let O be the event “the marble is odd-numbered” and B be the event “the marble is blue.” We want to find P(O|B).

    We can solve this by considering the reduced sample space. The condition “the marble is blue” means we only need to consider the 8 blue marbles.

    Step 1: Identify the new sample space.
    The sample space is the set of blue marbles: {B1, B2, B3, B4, B5, B6, B7, B8}. The total number of outcomes in this new space is 8.

    Step 2: Find the number of favorable outcomes within the new sample space.
    The favorable outcomes are the odd-numbered marbles within the blue marbles, which are {B1, B3, B5, B7}. There are 4 favorable outcomes.

    Step 3: Calculate the probability.
    P(O|B)=Number of odd blue marblesTotal number of blue marbles=48=12

    Answer (a): The probability is 1/2.

    b. Blue given that the marble is odd-numbered.

    Let B be the event “the marble is blue” and O be the event “the marble is odd-numbered.” We want to find P(B|O).

    The condition “the marble is odd-numbered” reduces our sample space.

    Step 1: Identify the new sample space.
    The odd-numbered marbles are:

    Red: {R1, R3} (2 marbles)

    Blue: {B1, B3, B5, B7} (4 marbles)
    The new sample space is all odd-numbered marbles, so the total number of outcomes is 2 + 4 = 6.

    Step 2: Find the number of favorable outcomes within the new sample space.
    The favorable outcomes are the blue marbles within this group of odd-numbered marbles. There are 4 favorable outcomes.

    Step 3: Calculate the probability.
    P(B|O)=Number of blue odd marblesTotal number of odd marbles=46=23

    Answer (b): The probability is 2/3.

  25. Compute the probability of flipping a fair coin and getting heads, given that the previous flip was tails.
    Answer (click to Show/Hide)

    To solve this formally, we will define the sample space for two consecutive coin flips and use the definition of conditional probability.

    Let H be heads and T be tails. The sample space S for two flips has four equally likely outcomes:
    S = {HH, HT, TH, TT}

    Let A be the event that the second toss is heads, and B be the event that the first toss is tails. We want to find the conditional probability P(A|B).

    The formula for conditional probability is:
    P(A|B)=P(AB)P(B)

    Step 1: Find the probability of event B (the first toss is tails).
    The outcomes where the first toss is tails are {TH, TT}. There are 2 such outcomes out of 4 total.
    P(B)=24=12

    Step 2: Find the probability of the intersection of A and B (the first toss is tails AND the second is heads).
    The only outcome that satisfies both conditions is {TH}. There is 1 such outcome out of 4 total.
    P(AB)=14

    Step 3: Calculate the conditional probability.
    Now we substitute the values from the previous steps into the formula.
    P(A|B)=1/42/4=12

    This confirms that the outcome of the previous flip does not affect the outcome of the current flip, which is the definition of independence.

    Answer: The probability of getting heads is 1/2.

  26. Find the probability of rolling a “1” on a fair die, given that the last 3 rolls were all ones.
    Answer (click to Show/Hide)

    This problem involves independent events. The outcome of one roll of a fair die has no effect on the outcome of any subsequent rolls. Each roll is a separate, independent event.

    Let A be the event “the current roll is a 1,” and let B be the event “the last 3 rolls were all ones.” We want to find the conditional probability P(A|B).

    Step 1: Understand Independent Events.
    Two events are independent if the occurrence of one does not affect the probability of the other. For independent events, the following is true:
    P(A|B)=P(A)
    The information that the last 3 rolls were all ones is irrelevant to the outcome of the current roll.

    Step 2: Calculate the probability of rolling a “1”.
    For a single, fair six-sided die, there are six equally likely outcomes {1, 2, 3, 4, 5, 6}. The probability of rolling a “1” is:
    P(A)=16

    Because the events are independent, the condition that the previous three rolls were ones does not change this probability.

    Answer: The probability of rolling a “1” given that the last 3 rolls were a one is 1/6.

  27. Suppose a math class contains 25 students, 14 females (three of whom speak French) and 11 males (two of whom speak French). Compute the probability that a randomly selected student speaks French, given that the student is female.
    Answer (click to Show/Hide)

    This is a conditional probability problem. We are given a condition—that the selected student is female—which reduces our total sample space.

    Let S be the event “the student speaks French,” and let F be the event “the student is female.” We want to find the probability P(S|F).

    Step 1: Determine the reduced sample space.
    The condition “given that the student is female” means we are only considering the 14 female students as our possible outcomes.

    Total outcomes in the new sample space = 14.

    Step 2: Determine the number of favorable outcomes within the new sample space.
    A favorable outcome is a student who speaks French. The problem states that within the group of 14 females, three speak French.

    Number of favorable outcomes = 3.

    Step 3: Calculate the probability.
    The probability is the ratio of the favorable outcomes to the total outcomes in our reduced sample space.
    P(Speaks French|Is Female)=Number of females who speak FrenchTotal number of females=314

    Answer: The probability that a randomly selected student speaks French, given that the student is female, is 3/14.

  28. Suppose a math class contains 25 students, 14 females (three of whom speak French) and 11 males (two of whom speak French). Compute the probability that a randomly selected student is male, given that the student speaks French.
    Answer (click to Show/Hide)

    This is a conditional probability problem. We are given a condition—that the selected student speaks French—which reduces our total sample space.

    Let M be the event “the student is male,” and let S be the event “the student speaks French.” We want to find the probability P(M|S).

    Step 1: Determine the reduced sample space.
    The condition “given that the student speaks French” means we are only considering the students who speak French as our possible outcomes.

    Number of females who speak French = 3

    Number of males who speak French = 2
    The total number of students who speak French is 3 + 2 = 5.

    Total outcomes in the new sample space = 5.

    Step 2: Determine the number of favorable outcomes within the new sample space.
    A favorable outcome is a student who is male. The problem states that within the group of French speakers, two are male.

    Number of favorable outcomes = 2.

    Step 3: Calculate the probability.
    The probability is the ratio of the favorable outcomes to the total outcomes in our reduced sample space.
    P(Is Male|Speaks French)=Number of males who speak FrenchTotal number of students who speak French=25

    Answer: The probability that a randomly selected student is male, given that the student speaks French, is 2/5.

  29. Forty eight percent of all undergraduates at a large university are male, 18% are majoring in business, and 10% are male and are majoring in business. Let M be the event that a randomly selected undergraduate is male and let B be the event that a randomly selected undergraduate is majoring in business.
    1. Compute P(M|B) and P(B|M). Interpret these results (include what the value is and what it represents in terms of the application).
    2. Are M and B independent? (justify your answer using the definition of independence from this section)
    Answer (click to Show/Hide)

    From the problem statement, we are given the following probabilities:

    P(M)=0.48

    P(B)=0.18

    P(MB)=0.10

    a. Compute and interpret P(M|B) and P(B|M).

    First, we compute P(M|B).
    This is the probability that a student is male, given that they are a business major.
    P(M|B)=P(MB)P(B)=0.100.180.556
    Interpretation: About 55.6% of business majors at this university are male.

    Next, we compute P(B|M).
    This is the probability that a student is a business major, given that they are male.
    P(B|M)=P(MB)P(M)=0.100.480.208
    Interpretation: About 20.8% of male undergraduates at this university are majoring in business.

    Answer (a): P(M|B)0.556 and P(B|M)0.208.

    b. Are M and B independent?

    Two events M and B are independent if P(M|B)=P(M). We can check if this condition is met using the values we have.

    From our calculation in part (a), P(M|B)0.556.

    We are given that P(M)=0.48.

    Since 0.5560.48, the events are not independent.

    Answer (b): The events M and B are not independent because the probability that a student is male changes depending on whether or not they are a business major.

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