2.3 Working with Events

Robert Foth

2.3 Working with Events

2.3: Working with Events

Learning Objectives

Upon completion of this section, you should be able to

Solve probability applications using the complement rule.
Find the probability when there is a union of two events.
Find the probability when there is an intersection of two events.

Complement of an Event

Now let us examine the probability that an event does not happen. Let’s start with a simple example: rolling a six-sided fair die. We already saw that the probability of rolling a six is: P(six) = 1/6. But what about not rolling a six? There are 5 outcomes that are not a six, so this give us: $P (not a six) = \frac{5}{6}$ .

Notice something interesting here: $P (six) + P (not a six) = \frac{1}{6} + \frac{5}{6} = \frac{6}{6} = 1$

This isn’t just a coincidence. When we consider an event “E” and its opposite “Not E“, these two events cover all outcomes in the sample space. In addition these two events do not share any of the outcomes as they are opposites of each other (either an outcome is in E or it is not E). The sum of their probabilities must then add up to 1 as there are no other collection of outcomes left from the sample space.

Complement of an Event

The complement of an event is the event “not E

The notation to represent the complement in this text will be “not E” or $E^{C}$

We can compute the probability of the complement using $P (E^{C}) = 1 - P (E)$

Notice that $P (E) = 1 - P (E^{C})$ and $1 = P (E) + P (E^{C})$ is also true.

Example 1

If you pull a random card from a deck of playing cards, what is the probability it is not a heart?

Solution

There are 13 hearts in the deck, so $P (heart) = \frac{13}{52} = \frac{1}{4}$ .

The probability of not drawing a heart is the complement: $P (not heart) = 1 - P (heart) = 1 - \frac{1}{4} = \frac{3}{4}$

Video Solution Example (1 min 40 secs – CC)

Try it Now 1

If the weather forecast predicts there is a 20% chance of rain, then what is the percent chance of it not raining?

Answer (click to Show/Hide)

The event “not raining” is the complement for the event raining, so the percent chance of not raining is
found to be:

$\begin{array}{l} P (R^{C}) = 1 - P (R) \\ P (R^{C}) = 1 - 0.20 \\ P (R^{C}) = 0.80 \end{array}$

We can conclude that there is a 80% chance of it not raining.

Intersection of events

As we continue our study in probability we need a way to describe what outcomes two events have in common. We can do this by bringing in new notation (as shown below) and call it an intersection.

Intersection of Events

The intersection of events A and B, denoted $A \cap B$ , is the collection of all outcomes that are in common to both events A and B. It corresponds to combining descriptions of the two events using the word “and.”

To say that the event $A \cap B$ occurred means that on a particular trial of the experiment both A and B occurred. A visual representation of the intersection of events A and B in a sample space S is given below “The Intersection of Events “. The intersection corresponds to the shaded lens-shaped region that lies within both ovals.

Sample space with events A and B shaped as ovals overlapping and the overlap shaded in — The Intersection of Events

Example 2

In the experiment of rolling a single die, find the intersection $E \cap T$ of the events

E: “the number rolled is even”
T: “the number rolled is greater than two.”

Solution

The sample space is $S = {1, 2, 3, 4, 5, 6}$ .

The outcomes in $E = {2, 4, 6}$ and $T = {3, 4, 5, 6}$ .

The outcomes that E and T have in common are 4 and 6, so $E \cap T = {4, 6}$ .

In words the intersection is described by “the number rolled is even and is greater than two.” The only numbers between one and six that are both even and greater than two are four and six, corresponding to $E \cap T$ given above.

Example 3

A single die is rolled:

Suppose the die is fair. Find the probability that the number rolled is both even and greater than two.
Suppose the die has been “loaded” so that $P (1) = \frac{1}{12}$ , $P (6) = \frac{3}{12}$ , and the remaining four outcomes are equally likely with one another. Now find the probability that the number rolled is both even and greater than two.

Solution

In both cases the sample space is $S = {1, 2, 3, 4, 5, 6}$ and the event in question is the intersection $E \cap T = {4, 6}$ of the previous example.

Since the die is fair, all outcomes are equally likely, so by counting we have $P (E \cap T) = \frac{2}{6}$
The information on the probabilities of the six outcomes that we have so far is:

Outcomes 1 2 3 4 5 6

Probability $\frac{1}{12}$ p p p p $\frac{3}{12}$

Since $P (1) + P (2) + P (3) + P (4) + P (5) + P (6) = 1$ , we can solve for what p is:

$\begin{aligned} \frac{1}{12} + P (2) + P (3) + P (4) + P (5) + \frac{3}{12} & = 1 \\ \frac{1}{12} + p + p + p + p + \frac{3}{12} & = 1 \\ \frac{4}{12} + 4 p & = 1 \\ 4 p & = 1 - \frac{4}{12} \\ 4 p & = \frac{12}{12} - \frac{4}{12} \\ 4 p & = \frac{8}{12} \\ p & = \frac{8}{12} \cdot \frac{1}{4} \\ p & = \frac{2}{12} \end{aligned}$

Thus we now have $P (4) = \frac{2}{12}$ and we can now find $P (E \cap T)$ :
$\begin{array}{l} P (E \cap T) = P (4) + P (6) \\ P (E \cap T) = \frac{2}{12} + \frac{3}{12} \\ P (E \cap T) = \frac{5}{12} \end{array}$

In some cases when we look at an intersection there may be no common outcomes between the two events. For A and B to have no outcomes in common means precisely that it is impossible for both A and B to occur on a single trial of the random experiment. This gives the following rule.

Mutually Exclusive Events

Two or more events are mutually exclusive if they cannot happen at the same time in a single trial of an experiment. This means that if one of these events occurs, it’s impossible for any of the others to occur simultaneously.

Events A and B are mutually exclusive if and only if $P (A \cap B) = 0$ .

Example 4

Determine if the events are mutually exclusive for the the experiment of randomly selecting a card from a deck of standard playing cards.

D is the event that you select a diamond and R is the event you select a red card.
D is the event that you select a diamond and B is the event you select a black card.
Q is the event that you select a Queen and R is the event you select a red card.

Solution

The diamonds are already red cards, so the events D and R share in common all the cards that are diamonds, so they are not mutually exclusive.
The diamond cards are red, so there are no outcomes in D that are also in the outcome that the card is black. We would say D and B are mutually exclusive and that $P (D \cap B) = 0$ .
There are two black queens and two red queens, so with the event Q we have two outcomes that are in common with the event R that we select a red card. These two events are not mutually exclusive.

Union of Events

Now that we have a way to talk about what outcomes two events have in common we will look at how we describe the total outcomes in two or more events. These outcomes do not have to sit within both (or many) events to be part of that group.

Union of Events

The union of events A and B, denoted $A \cup B$ , is the collection of all outcomes that are in events A and B, or of both of them. It corresponds to combining descriptions of the two events using the word “or.”

To say that the event $A \cup B$ occurred means that on a particular trial of the experiment either A or B occurred (or both did). A visual representation of the union of events A and B in a sample space S is given in “The Union of Events.” The union corresponds to the shaded region.

Sample space with events A and B shaped as ovals overlapping and the both ovals shaded in the same color — The Union of Events

Example 5

In the experiment of rolling a single die, find the union of the events

E: “the number rolled is even”
T: “the number rolled is greater than two.”

Solution

The sample space is $S = {1, 2, 3, 4, 5, 6}$ .

The outcomes in $E = {2, 4, 6}$ and $T = {3, 4, 5, 6}$ .

The union E and T is made up of all outcomes they have together: $E \cup T = {2, 3, 4, 5, 6}$ . Note that an outcome such as 4 that is in both sets is still listed only once (although strictly speaking it is not incorrect to list it twice).

In words the union is described by “the number rolled is even or is greater than two.” Every number between one and six except the number one is either even or is greater than two, corresponding to $E \cup T$ given above.

The following Additive Rule of Probability is a useful formula for calculating the probability of $A \cup B$ .

Addition Rule

The additive rule of probability states how to calculate the probability of the union of two events. For any two events the probability of A or B is:

$\begin{array}{l} P (A or B) = P (A) + P (B) - P (A and B) \\ P (A \cup B) = P (A) + P (B) - P (A \cap B) \end{array}$

The next example, in which we compute the probability of a union both by counting and by using the formula, shows why the last term in the formula is needed.

Example 6

Suppose we flipped a coin and rolled a die, and wanted to know the probability of getting a head on the coin or a 6 on the die.

Solution

Here, there are still 12 possible outcomes: S = {H1,H2,H3,H4,H5,H6,T1,T2,T3,T4,T5,T6}

By simply counting, we can see that 7 of the outcomes have a head on the coin or a 6 on the die or both – we use or inclusively here (these 7 outcomes are H1, H2, H3, H4, H5, H6, T6), so the probability is $\frac{7}{12}$ .

How could we have found this from the individual probabilities of tossing a coin and observing Heads and rolling a die and observing a 6?

As we would expect, $\frac{1}{2}$ of these outcomes have a head, and $\frac{1}{6}$ of these outcomes have a 6 on the die. If we add these, $\frac{1}{2} + \frac{1}{6} = \frac{6}{12} + \frac{2}{12} = \frac{8}{12}$ , which is not the correct probability. Looking at the outcomes we can see why: the outcome H6 would have been counted twice, since it contains both a head and a 6; the probability of both a head and rolling a 6 is $\frac{1}{12}$ .

If we subtract out this double count, we have the correct probability: $\frac{8}{12} - \frac{1}{12} = \frac{7}{12}$ .

Putting the above into the formula we see the formula for the Additive Rule in work:

$\begin{array}{l} P (Heads or roll a 6) & = P (Heads) + P (roll a 6) - P (Heads and roll a 6) \\ = \frac{1}{2} + \frac{1}{6} - \frac{1}{12} \\ = \frac{6}{12} + \frac{2}{12} - \frac{1}{12} \\ = \frac{8}{12} - \frac{1}{12} \\ = \frac{7}{12} \end{array}$

Video Solution (2 min 27 secs – CC) Another example starts after the 2:27 mark of the video.

Example 7

Two fair dice are thrown. Find the probabilities of the following events:

both dice show a four
at least one die shows a four (which can be translated as the first die is a four or the second die is a four)

Solution

As was the case with tossing two identical coins, actual experience dictates that for the sample space to have equally likely outcomes we should list outcomes as if we could distinguish the two dice. We could imagine that one of them is red and the other is green. Then any outcome can be labeled as a pair of numbers as in the following display, where the first number in the pair is the number of dots on the top face of the green die and the second number in the pair is the number of dots on the top face of the red die.

$\begin{matrix} (1, 1) & (1, 2) & (1, 3) & (1, 4) & (1, 5) & (1, 6) \\ (2, 1) & (2, 2) & (2, 3) & (2, 4) & (2, 5) & (2, 6) \\ (3, 1) & (3, 2) & (3, 3) & (3, 4) & (3, 5) & (3, 6) \\ (4, 1) & (4, 2) & (4, 3) & (4, 4) & (4, 5) & (4, 6) \\ (5, 1) & (5, 2) & (5, 3) & (5, 4) & (5, 5) & (5, 6) \\ (6, 1) & (6, 2) & (6, 3) & (6, 4) & (6, 5) & (6, 6) \end{matrix}$

There are 36 equally likely outcomes, of which exactly one corresponds to two fours, so the probability of a pair of fours is $\frac{1}{36}$ .
First Approach: Counting. From the table we can see that there are 11 pairs that correspond to the event in question: the six pairs in the fourth row (the first die shows a four) plus the additional five pairs other than the pair (4,4), already counted, in the fourth column (the second die is afour), so the answer is $\frac{11}{36}$
Second Approach: Using the Additive Rule. Let A be the event the first roll is a 4 and B be the event that the second roll is a 4. Then let us start with first writing out the additive rule in this case:

$P (A or B) = P (A) + P (B) - P (A and B)$

There are six outcomes for the event A as we can see in row 4 where the first dice rolled is a 4. There are also six outcomes for event B as we can see in column 4 where the second dice rolled is a 4. Finally there is only one outcome where both the first and the second dice rolled is a four (where the 4th row and 4th column intersect each other). The additive rule would have us find the probabilty as:

$P (A or B) = \frac{6}{36} + \frac{6}{36} - \frac{1}{36} = \frac{11}{36}$

The first two terms in the additive rule is adding up the total number of outcomes in the two events, but it ignores any duplication so we end up double counting any outcomes that they have in common. The last term takes care to remove the double counts, so that we end up with the correct result.

Example 8

Suppose we draw one card from a standard deck.

What is the probability that we get a Queen or a King?
What is the probability that we get a red card or a King?

Solution

There are 4 Queens and 4 Kings in the deck, hence 8 outcomes corresponding to a Queen or King out of 52 possible outcomes. Thus the probability of drawing a Queen or a King is:
$P (King or Queen) = \frac{8}{52}$

Note that in this case, there are no cards that are both a Queen and a King, so $P (King and Queen) = 0$ . Using our additive probability rule, we could have said:

$\begin{array}{l} P (King or Queen) = P (King) + P (Queen) - P (King and Queen) \\ P (King or Queen) = \frac{4}{52} + \frac{4}{52} - 0 \\ P (King or Queen) = \frac{8}{52} \end{array}$
Half the cards are red, so $P (Red) = \frac{26}{52}$
There are four kings, so $P (King) = \frac{4}{52}$

There are two red kings, so $P (Red and King) = \frac{2}{52}$

We can then calculate
$\begin{array}{l} P (Red or King) = P (Red) + P (King) - P (Red and King) \\ P (Red or King) = \frac{26}{52} + \frac{4}{52} - \frac{2}{52} \\ P (Red or King) = \frac{28}{52} \end{array}$

Video Solution (2 min 21 secs – CC)

Try it Now 2

A tutoring service specializes in preparing adults for high school equivalence tests. Among all the students seeking help from the service, 63% need help in mathematics, 34% need help in English, and 27% need help in both mathematics and English. What is the percentage of students who need help in either mathematics or English?

Hint 1 (click to Show/Hide)

Imagine selecting a student at random, that is, in such a way that every student has the same chance of being selected. Let M denote the event “the student needs help in mathematics” and let E denote the event “the student needs help in English.” Now write out all the given information into probability values using the probability notation.

63% need help in mathematics: $P (M) = 0.63$
34% need help in English: $P (E) = 0.34$
27% need help in both mathematics and English: $P (M and E) = 0.27$

Next write out the question in probability notation and see if there is something you can use to solve.

Answer (click to Show/Hide)

The given question is asking for $P (M or E)$ . With the given information (below) use the additive rule to find a solution.

63% need help in mathematics: $P (M) = 0.63$
34% need help in English: $P (E) = 0.34$
27% need help in both mathematics and English: $P (M and E) = 0.27$

Now use the Additive Rule for Probability

$\begin{array}{l} P (M or E) = P (M) + P (E) - P (M and E) \\ P (M or E) = 0.63 + 0.34 - 0.27 \\ P (M or E) = 0.70 \end{array}$

The probability that a randomly selected student is selected and they need help in either math or english was found to be 0.70. We could also say 70% of the students need help in either math or english.

Note how the reasoning that if 63% need help in mathematics and 34% need help in English then 63 plus 34 or 97% need help in one or the other gives a number that is too large. The percentage that need help in both subjects must be subtracted off, else the people needing help in both are counted twice, once for needing help in mathematics and once again for needing help in English. The simple sum of the probabilities would work if the events in question were mutually exclusive, for then $P (A \cap B)$ is zero, and makes no difference.

We will finish this section with an example that shows data given to us in a two-way classification table.

Example 9

Volunteers for a disaster relief effort were classified according to both specialty (C: construction, E: education, M: medicine) and language ability (S: speaks a single language fluently, T: speaks two or more languages fluently). The results are shown in the following two-way classification table:

Specialty	Language Ability
Specialty	S	T
C	12	1
E	4	3
M	6	2

The first row of numbers means that 12 volunteers whose specialty is construction speak a single language fluently, and 1 volunteer whose specialty is construction speaks at least two languages fluently. Similarly for the other two rows.

A volunteer is selected at random, meaning that each one has an equal chance of being chosen. Find the probability that:

their specialty is medicine and they speak two or more languages;
either their specialty is medicine or they speak two or more languages;
their specialty is something other than medicine.

Solution

When information is presented in a two-way classification table it is typically convenient to adjoin to the table the row and column totals, to produce a new table like this:

Specialty	Language Ability		Total
Specialty	S	T	Total
C	12	1	13
E	4	3	7
M	6	2	8
Total	22	6	28

The probability sought is $P (M \cap T)$ . The table shows that there are 2 such people, out of 28 in all, hence $P (M \cap T) = \frac{2}{28} \approx 0.07$ or about a 7% chance.
The probability sought is $P (M \cup T)$ . The third row total and the grand total in the sample give $P (M) = \frac{2}{28}$ . The second column total and the grand total give $P (T) = \frac{6}{28}$ .
$\begin{array}{l} P (M \cup T) = P (M) + P (T) - P (M \cap T) \\ P (M \cup T) = \frac{8}{28} + \frac{6}{28} - \frac{2}{28} \\ P (M \cup T) = \frac{12}{28} \\ P (M \cup T) \approx 0.43 \end{array}$
Thus using the result from part (a) is about a 43% chance.
This probability can be computed in two ways. Since the event of interest can be viewed as the event $C \cup E$ and the events C and E are mutually exclusive, the answer is, using the first two row totals,
$\begin{array}{l} P (C \cup E) = P (C) + P (E) - P (C \cap E) \\ P (C \cup E) = \frac{13}{28} + \frac{7}{28} - \frac{0}{28} \\ P (C \cup E) = \frac{20}{28} \\ P (C \cup E) \approx 0.71 \end{array}$

On the other hand, the event of interest can be thought of as the complement $M^{C}$ of M, hence using the value of $P (M)$ computed in part (b),

$\begin{array}{l} P (M^{C}) = 1 - P (M) \\ P (M^{C}) = 1 - \frac{8}{28} \\ P (M^{C}) = \frac{20}{28} \\ P (M^{C}) \approx 0.71 \end{array}$ which gives us the same value as before.

Exercises

For the sample space S={a,b,c,d,e} identify the complement of each event given.
1. $A = {a, d, e}$
2. $B = {b, c, d, e}$
3. S
Answer (click to Show/Hide)
1. $A^{C} = {b, c}$
2. $B^{C} = {a}$
3. $S^{C} = \emptyset$ : This is called the empty set as the complement of the entire smaple space consists of no outcomes.
For the sample space S={r,s,t,u,v} identify the complement of each event given.
1. $R = {t, u}$
2. $T = {r}$
3. $\emptyset$ (the “empty” set that has no elements)
Answer (click to Show/Hide)
1. $R^{C} = {r, s, v}$
2. $T^{C} = {s, t, u, v}$
3. The complement would be the entire sample space S
The sample space for three tosses of a coin is
S={hhh,hht,hth,thh,htt,tht,tth,ttt}
The events H and M are defined below.
H : at least one head is observed
M: more heads than tails are observed

Complete the following
1. List the outcomes that comprise H and all the outcomes that comprise M.
2. List the outcomes that comprise $H \cap M$ , $H \cup M$ , and $H^{C}$ .
3. Assuming all outcomes are equally likely, find $P (H \cap M)$ , $P (H \cup M)$ , and $P (H^{C})$ .
4. Determine whether or not $P (H^{C})$ and M are mutually exclusive. Explain why or why not.
Answer (click to Show/Hide)
1. $H = {h h h, h h t, h t h, t h h, h t t, t h t, t t h}$ , $M = {h h h, h h t, h t h, t h h}$
2. $H \cap M = {h h h, h h t, h t h, t h h}$ ,
  $H \cup M = {h h h, h h t, h t h, t h h, h t t, t h t, t t h} = H$ ,
  
  $H^{C} = {t t t}$
3. $P (H \cap M) = \frac{4}{8}$ ,
  $P (H \cup M) = \frac{7}{8}$ ,
  $P (H^{C}) = \frac{1}{8}$
4. Mutually exclusive because they have no elements in common.
For the experiment of rolling a single six-sided die once the events T and G are defined below.
T: the number rolled is three
G: the number rolled is four or greater
Complete the following
1. List the outcomes that comprise T and G.
2. List the outcomes that comprise $T \cap G$ , $T \cup G$ , $T^{C}$ , and ${(T \cup G)}^{C}$ .
3. Assuming all outcomes are equally likely, find $P (T \cap G)$ , $P (T \cup G)$ , $P ({(T \cup G)}^{C})$ , and $P (T^{C})$ .
4. Determine whether or not T and G are mutually exclusive. Explain why or why not.
Answer (click to Show/Hide)
1. $T = {3}$ and $G = {4, 5, 6}$
2. The outcomes are:
  $\begin{array}{l} T \cap G = \emptyset \\ T \cup G = {3, 4, 5, 6} \\ T^{C} = {1, 2, 4, 5, 6} \\ {(T \cup G)}^{C} = {1, 2} \end{array}$
3. The probabilities are:
  $\begin{array}{l} P (T \cap G) = 0 \\ P (T \cup G) = \frac{4}{6} \\ P (T^{C}) = 1 - P (T) = 1 - \frac{1}{6} = \frac{5}{6} \\ P ({(T \cup G)}^{C}) = 1 - P (T \cup G) = 1 - \frac{4}{6} = \frac{2}{6} \end{array}$
4. They are mutually exclusive as they share not outcome in common. T consists of only the outcome of 3 and we do not find that in the set of outcomes for G.
Confirm that the probabilities in the two-way contingency table add up to 1, then use it to find the probabilities of the events indicated.

U V W

A 0.15 0.00 0.23

B 0.22 0.30 0.10
1. $P (A)$ , $P (B)$ , $P (A \cap B)$ .
2. $P (U)$ , $P (W)$ , $P (U \cap W)$ .
3. $P (U \cup W)$ .
4. $P (V^{C})$ .
5. Determine whether or not the events A and U are mutually exclusive; the events A and V.
Answer (click to Show/Hide)
1. $P (A) = 0.38$ , $P (B) = 0.62$ , $P (A \cap B) = 0$
2. $P (U) = 0.37$ , $P (W) = 0.33$ , $P (U \cap W) = 0$
3. The probability is:
  $\begin{array}{l} P (U \cup W) = P (U) + P (W) - P (U \cap W) \\ P (U \cup W) = 0.37 + 0.33 + 0 \\ P (U \cup W) = 0.70 \end{array}$
4. $P (V^{C}) = 1 - P (V) = 1 - 0.30 = 0.70$ , take note how the complement of V is all of U or W.
5. $P (A \cap U) = 0.15 \neq 0$ , so A and U are not mutually exclusive.
  $P (A \cap V) = 0$ , A and V are mutually exclusive.
Make a statement in ordinary English that describes the complement of each event (do not simply insert the word “not”).
1. In the roll of a die: “five or more.”
2. In a roll of a die: “an even number.”
3. In two tosses of a coin: “at least one heads.”
4. In the random selection of a college student: “Not a freshman.”
Answer (click to Show/Hide)
1. Roll a four or less.
2. Roll an odd number.
3. No heads
4. Select a freshman
A tourist who speaks English and German but no other language visits a region of Slovenia. If 35% of the residents speak English, 15% speak German, and 3% speak both English and German, what is the probability that the tourist will be able to talk with a randomly encountered resident of the region?

Answer (click to Show/Hide)

The randomly chosen resident must speak English or German (or both).

$\begin{array}{l} P (E \cup G) = P (E) + P (G) - P (E \cap G) \\ P (E \cup G) = 0.35 + 0.15 - 0.03 \\ P (E \cup G) = 0.47 \end{array}$
The table relates the response to a fund-raising appeal by a college to its alumni to the number of years since graduation.

Response Years Since Graduation

0–5 6–20 21–35 Over 35

Responded 120 440 210 90

No Response 1380 3560 3290 910

An alumnus is selected at random. Adjoin the row and column totals to the table and use the expanded table to find the probability of each of the following events.
1. The alumnus responded.
2. The alumnus did not respond.
3. The alumnus graduated at least 21 years ago.
4. The alumnus graduated at least 21 years ago and responded.
Answer (click to Show/Hide)
1. $P (R) = \frac{120 + 440 + 210 + 90}{120 + 440 + 210 + 90 + 1380 + 3560 + 3290 + 910} = \frac{860}{10000}$
2. $P (R^{C}) = 1 - P (R) = 1 - \frac{860}{10000} = \frac{9140}{10000}$
3. $P (at least 21 years) = \frac{210 + 90 + 3290 + 910}{10000} = \frac{4500}{10000}$
4. $P (at least 21 years \cap R) = \frac{210 + 90}{10000} = \frac{300}{10000}$
Compute the probability of rolling a fair 12-sided die (numbered 1 to 12) and getting a number other than 8.

Answer (click to Show/Hide)

$P (not 8) = 1 - P (roll a 8) = 1 - \frac{1}{12} = \frac{11}{12}$
Referring to the grade table below, what is the probability that a student chosen at random did NOT earn a C?

A B C Total

Male 8 18 13 39

Female 10 4 12 26

Total 18 22 25 65

Answer (click to Show/Hide)

$P (Not a C) = 1 - P (C) = 1 - \frac{25}{65} = \frac{40}{65}$
Referring to the credit card table below, what is the probability that a person chosen at random has at least one credit card? Answer the question with and without using the complement rule.

Zero One Two or more Total

Male 9 5 19 33

Female 18 10 20 48

Total 27 15 39 81

Answer (click to Show/Hide)

Without Complement Rule:

$P (at least 1 CC) = \frac{15 + 39}{81} = \frac{54}{81}$ With Complement Rule:

$P (at least 1 CC) = 1 - P (No CC) = 1 - \frac{27}{81} = \frac{54}{81}$
A fair die is rolled twice. Find the follow probabilities:
1. Showing a 5 on the first roll and an even number on the second roll?
2. Showing a 5 on the first roll or an even number on the second roll?
Answer (click to Show/Hide)
Let F1 be the event of an even on the first roll and E2 be the event of even on second roll
1. $P (F1 and E2) = \frac{3}{36}$
2. Use the additive rule:
  $\begin{array}{l} P (F1 or E2) = P (F 1) + P (E 2) - P (F 1 and E2) \\ P (F1 or E2) = \frac{6}{36} + \frac{18}{36} - \frac{3}{36} \\ P (F1 or E2) = \frac{21}{36} \end{array}$
Suppose that P(E)=0.4,P(F)=0.5. From the limited information, find a maximum and minimum value for P(E or F). Make sure to explain your answer for each case.
Answer (click to Show/Hide)
1. Maximum: 0.90. E and F are mutually exclusive.
2. Minimum: 0.50. All outcomes in F are also in E.
Suppose that P(E)=0.4,P(F)=0.5. From the limited information, find a maximum and minimum value for P(E and F). Make sure to explain your answer for each case.
Answer (click to Show/Hide)
1. Maximum: 0.40. All outcomes in F are also in E, but largest intersection can be is F.
2. Minimum: 0. F and E are mutually exclusive.
The probability that you get a raise is 0.68, the probability that you get your own office is 0.41, and the probability that you get a raise and your own office is 0.24. Answer the following:
1. What is the probability you get a raise or your own office?
2. What is the probability you do not get your own office?
Answer (click to Show/Hide)
Let R be the event you get a raise and O be the event you get your own office.
1. Use the additive rule.
  $\begin{array}{l} P (R or O) = P (R) + P (O) - P (R and O) \\ P (R or O) = 0.68 + 0.41 - 0.24 \\ P (R or O) = 0.85 \end{array}$
2. $P (O^{C}) = 1 - P (O) = 1 - 0.41 = 0.59$

Attributions

This page contains modified content from David Lippman, “Math In Society, 2nd Edition.” Licensed under CC BY-SA 4.0.
This page contains modified content from “College Algebra 2017.” Lumen Learning. Licensed under CC BY 4.0.
This page contains modified content from Introductory Statistics Copyright © 2024 by LOUIS: The Louisiana Library Network is licensed under a Creative Commons Attribution-ShareAlike 4.0 International License, except where otherwise noted.
This page contains content by Robert Foth, Math Faculty, Pima Community College, 2021. Licensed under CC BY 4.0.

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Response	Years Since Graduation
Response	0–5	6–20	21–35	Over 35
Responded	120	440	210	90
No Response	1380	3560	3290	910

Outcomes	1	2	3	4	5	6
Probability	$\frac{1}{12}$	p	p	p	p	$\frac{3}{12}$

	U	V	W
A	0.15	0.00	0.23
B	0.22	0.30	0.10

	A	B	C	Total
Male	8	18	13	39
Female	10	4	12	26
Total	18	22	25	65

	Zero	One	Two or more	Total
Male	9	5	19	33
Female	18	10	20	48
Total	27	15	39	81

Complement of an Event

Solution

Intersection of events

Solution

Solution

Solution

Union of Events

Solution

Solution

Solution

Solution

Solution

Exercises

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