4.3: The Addition and Multiplication Rules of Probability

Introduction

When calculating probability, we can always approach the questions intuitively as demonstrated in the previous section, but there are formulas that can simplify the process for probabilities about compound events where we have an event that could have multiple outcomes and also probabilities for more than one simple event occurring. Additionally, when we don't know the sample space, but instead the probability of an outcome, then we can still calculate the probabilities for compound and multiple events. For these formulas to work, we need keep in mind whether two events are independent or dependent and if they are mutually exclusive or not.

The Multiplication and Addition Rules

Definition: Multiplication Rule

If \(A\) and \(B\) are two events defined on a sample space, then:

\[P(\text{A AND B}) = P(B)P(A|B) \label{eq1}\]

This rule may also be written as:

\[P(A|B) = \dfrac{P(\text{A AND B})}{P(B)} \nonumber\]

(The probability of \(A\) given \(B\) equals the probability of \(A\) and \(B\) divided by the probability of \(B\).)

If \(A\) and \(B\) are independent, then

\[P(A|B) = P(A). \nonumber\]

and Equation \ref{eq1} becomes

\[P(\text{A AND B}) = P(A)P(B). \nonumber\]

 

Definition: Addition Rule

If \(A\) and \(B\) are defined on a sample space, then:

\[P(\text{A OR B}) = P(A) + P(B) - P(\text{A AND B}) \label{eq5}\]

If \(A\) and \(B\) are mutually exclusive, then

\[P(\text{A AND B}) = 0. \nonumber\]

and Equation \ref{eq5} becomes

\[P(\text{A OR B}) = P(A) + P(B). \nonumber\]

Note: \(P(\text{A AND  B}\) for the addition rule is just the overlap of events \(A\) and \(B\). In the event they are mutually exclusive, then there is no overlap, so there are 0 shared elements. Otherwise, if they aren't mutually exclusive then we don't want to double count their share elements when we add \(P(\text{A}\) and \(P(\text{B}\), so we subtract that overlap to adjust. 

Let's consider the case where Klaus is trying to choose where to go on vacation.

His two choices are: \(\text{A} = \text{New Zealand}\) and \(\text{B} = \text{Alaska}\) and he can't afford to go on two vacations. 

We know that the probability that he chooses \(\text{A}\) is \(P(\text{A}) = 0.6\) and the probability that he chooses \(\text{B}\) is \(P(\text{B}) = 0.35\).

\(P(\text{A AND B}) = 0\) because Klaus can only afford to take one vacation. This tells us events \(A\) and \(B\) are mutually exclusive.

Therefore, the probability that he chooses either New Zealand or Alaska is \(P(\text{A OR B}) = P(\text{A}) + P(\text{B}) = 0.6 + 0.35 = 0.95\).

Note that the probability that he does not choose to go to either location on vacation must be 0.05. Since probabilities sum 1 or 100% of the outcomes. 

\(P(\text{A|B}) = 0\) since if B happens then A cannot happen. The same is true if A happens, then B cannot happen. Since  \(P(\text{A|B}) \neq P(\text{A})\), the events are not independent. These are also called dependent events.

Example \(\PageIndex{1}\)

Carlos plays college soccer. He makes a goal 65% of the time he shoots. Carlos is going to attempt two goals in a row in the next game. Carlos tends to shoot in streaks. The probability that he makes the second goal GIVEN that he made the first goal is 0.90.

1. What is the probability that he makes both goals?
2. What is the probability that Carlos makes either the first goal or the second goal?
3. Are \(\text{A}\) and \(\text{B}\) independent?
4. Are \(\text{A}\) and \(\text{B}\) mutually exclusive?

Solutions

Let's pull the pertinent information from the question and assign some variables.

Let \(\text{A} =\) the event Carlos is successful on his first attempt, so \(P(\text{A}) = 0.65\).

Let \(\text{B} =\) the event Carlos is successful on his second attempt, so \(P(\text{B}) = 0.65\).

\(\text{B|A}\) is the event that Carlos makes his second goal GIVEN that he made the first goal, so \(P(\text{B|A}) = 0.90\)

a. The problem is asking you to find \(P(\text{A AND B})\). Recall that the order of the outcomes doesn't matter, so this could also be interpreted as P(\text{B AND A})\) since multiplication is commutative (meaning you can multiple in either order).

\(P(\text{A AND B}) = P(\text{A})P(\text{B|A}) = (0.65)(0.90) = 0.585\)

Carlos makes the first and second goals with probability 0.585.

b. The problem is asking you to find \(P(\text{A OR B})\). Since the case where he makes both goals counts for \(P(\text{A})\) and for \(P(\text{B})\), we want to subtract that value so we don't double count it. 

\[P(\text{A OR B}) = P(\text{A}) + P(\text{B}) - P(\text{A AND B}) = 0.65 + 0.65 - 0.585 = 0.715 \nonumber\]

Carlos makes either the first goal or the second goal with probability 0.715.

c. No, they are not, because \(P(\text{B | A}) \neq P(\text{B}) \), which means \(P(\text{B AND A})\) is not equal to \(P(\text{A})P(\text{B})\) .

Check multiplication rule:

\[P(\text{A})P(\text{B}) = (0.65)(0.65) = 0.423 \nonumber\]

\[P(\text{A})P(\text{B|A}) = (0.65)(0.90) = 0.585 \nonumber\]

\[0.423 \neq 0.585 \nonumber\]

d. No, they are not because \(P(\text{A AND B}) = 0.585\). To be mutually exclusive, \(P(\text{A AND B})\) must equal zero.

Exercise \(\PageIndex{1}\)

Helen plays basketball. For free throws, she makes the shot 75% of the time. Helen must now attempt two free throws. The probability that Helen makes the second free throw given that she made the first is 0.85. What is the probability that Helen makes both free throws?

Answer

 Pull the pertinent information from the question and assign some variables.

Let \(\text{C} =\) the event that Helen makes the first shot, so \(P(\text{C}) = 0.75\).

Let \(\text{D} =\) the event Helen makes the second shot, so \(P(\text{D}) = 0.75\).

\(\text{D|C}\) is the event that Helen makes her second shot GIVEN that she made the first shot, so \(P(\text{D|C}) = 0.85\)

We want to find \(P(\text{C AND D}) or P(\text{D AND C})\), remember the order you write these in is commutative just like multiplication.

\[P(\text{D|C}) = 0.85 \nonumber\]

\[P(\text{C AND D}) = P(\text{C})P(\text{D|C}) = (0.75)(0.85) = 0.6375 \nonumber\]

Helen makes the first and second free throws with probability 0.6375.

Example \(\PageIndex{2}\)

A community swim team has 150 members. Seventy-five of the members are advanced swimmers. Forty-seven of the members are intermediate swimmers. The remainder are novice swimmers. Forty of the advanced swimmers practice four times a week. Thirty of the intermediate swimmers practice four times a week. Ten of the novice swimmers practice four times a week. Suppose one member of the swim team is chosen randomly.

1. What is the probability that the member is a novice swimmer?
2. What is the probability that the member practices four times a week?
3. What is the probability that the member is an advanced swimmer and practices four times a week?
4. What is the probability that a member is an advanced swimmer and an intermediate swimmer? Are being an advanced swimmer and an intermediate swimmer mutually exclusive? Why or why not?
5. Are being a novice swimmer and practicing four times a week independent events? Why or why not?

Solutions

Let's pull the pertinent information from the question and assign some variables.

\(\text{A}\) is the event that we select a Advanced Swimmer. \(n(\text{A}) = 75\)

\(\text{B}\) is the event that we select an Intermediate Swimmer. \(n(\text{B}) = 47\)

\(\text{C}\) is the event that we select a Novice Swimmer. \(n(\text{C}) = ?\) We don't know this number just yet. 

\(\text{D}\) is the event that a swimmer practices four times a week. 

\(\text{D|A}\) is the even that a swimmer practices four times a week given they are advanced. \(n(\text{D|A}) = 40\)

\(\text{D|B}\) is the even that a swimmer practices four times a week given they are intermediate. \(n(\text{D|B}) = 30\)

\(\text{D|C}\) is the even that a swimmer practices four times a week given they are novice. \(n(\text{D|C}) = 10\)

Now that we have all of the information from the question organized, it's time to answer the questions. 

a. To find the probability that the selected member is a novice swimmer, we need to figure out how many novices there are out of the 150 swimmers. To do this, we will subtract the advanced and the intermediate swimmers from the total.

\(n(\text{C}) = 150 - 75 - 47 = 28\) so \(P((\text{C})= \dfrac{28}{150}\)

b. To find the probability of selecting a member who practices four times a week, we just need to add the provided numbers since they will either practice four times a week or not. There is no overlap. So, \(n(\text{D}) = 40 + 30+ 10 = 80\) and \(P(\text{D}) = \dfrac{80}{150}\)

c. To find the probability that we selected an advanced swimmer and they practice four times a week, we want

\[P(\text{A and D}) = P(\text{A})P(\text{D|A})\nonumber\]

\[P(\text{A and D}) = \frac{75}{150} \cdot \frac{40}{75} \nonumber\] since \(\text{D|A}\) means 40 out of 75 advanced swimmers practice 4 times a week.

\[P(\text{A and D})= \dfrac{40}{150} \nonumber\]

d. \(P(\text{advanced AND intermediate}) = 0\), so these are mutually exclusive events. A swimmer cannot be an advanced swimmer and an intermediate swimmer at the same time.

e. No, these are not independent events. Since \[P(\text{D|C}) \neq P(\text{D})\nonumber\]

\[P(\text{C AND D}) = 0.0667 \nonumber\]

\[P(\text{C})P(\text{D}) = 0.0996 \nonumber\]

\[0.0667 \neq 0.0996 \nonumber\]

Exercise \(\PageIndex{2}\)

Felicity attends Modesto JC in Modesto, CA. The probability that Felicity enrolls in a math class is 0.2 and the probability that she enrolls in a speech class is 0.65. The probability that she enrolls in a math class GIVEN that she enrolls in speech class is 0.25.

Let: \(\text{M} =\) math class, \(\text{S} =\) speech class, \(\text{M|S} =\) math given speech

1. What is the probability that Felicity enrolls in math and speech?
   Find \(P(\text{M AND S}) = P(\text{M|S})P(\text{S})\).
2. What is the probability that Felicity enrolls in math or speech classes?
   Find \(P(\text{M OR S}) = P(\text{M}) + P(\text{S}) - P(\text{M AND S})\).
3. Are \(\text{M}\) and \(\text{S}\) independent? Is \(P(\text{M|S}) = P(\text{M})\)?
4. Are \(\text{M}\) and \(\text{S}\) mutually exclusive? Is \(P(\text{M AND S}) = 0\)?

Answer

a. 0.1625, b. 0.6875, c. No, d. No

Exercise \(\PageIndex{3}\)

A student goes to the library. Let events \(\text{B} =\) the student checks out a book and \(\text{D} =\) the student check out a DVD. Suppose that \(P(\text{B}) = 0.40, P(\text{D}) = 0.30\) and \(P(\text{D|B}) = 0.5\).

1. Find \(P(\text{B AND D})\).
2. Find \(P(\text{B OR D})\).

Answer

1. \(P(\text{B AND D}) = P(\text{D|B})P(\text{B}) = (0.5)(0.4) = 0.20\).
2. \(P(\text{B OR D}) = P(\text{B}) + P(\text{D}) − P(\text{B AND D}) = 0.40 + 0.30 − 0.20 = 0.50\)

Exercise \(\PageIndex{4}\)

A student goes to the library. Let events \(\text{B} =\) the student checks out a book and \(\text{D} =\) the student checks out a DVD. Suppose that \(P(\text{B}) = 0.40, P(\text{D}) = 0.30\) and \(P(\text{D|B}) = 0.5\).

1. Find \(P(\text{B′})\).
2. Find \(P(\text{D AND B})\).
3. Find \(P(\text{B|D})\).
4. Find \(P(\text{D AND B′})\).
5. Find \(P(\text{D|B′})\).

Answer

1. \(P(\text{B′}) = 0.60\)
2. \(P(\text{D AND B}) = P(\text{D|B})P(\text{B}) = 0.20\)
3. \(P(\text{B|D}) = \dfrac{P(\text{B AND D})}{P(\text{D})} = \dfrac{(0.20)}{(0.30)} = 0.66\)
4. \(P(\text{D AND B′}) = P(\text{D}) - P(\text{D AND B}) = 0.30 - 0.20 = 0.10\)
5. \(P(\text{D|B′}) = P(\text{D AND B′})P(\text{B′}) = (P(\text{D}) - P(\text{D AND B}))(0.60) = (0.10)(0.60) = 0.06\)

Example \(\PageIndex{3}\)

Studies show that about one woman in seven (approximately 14.3%) who live to be 90 will develop breast cancer. Suppose that of those women who develop breast cancer, a test is negative 2% of the time. Also suppose that in the general population of women, the test for breast cancer is negative about 85% of the time. Let \(\text{B} =\) woman develops breast cancer and let \(\text{N} =\) tests negative. Suppose one woman is selected at random.

1. What is the probability that the woman develops breast cancer? What is the probability that woman tests negative?
2. Given that the woman has breast cancer, what is the probability that she tests negative?
3. What is the probability that the woman has breast cancer AND tests negative?
4. What is the probability that the woman has breast cancer or tests negative?
5. Are having breast cancer and testing negative independent events?
6. Are having breast cancer and testing negative mutually exclusive?

Solutions

1. \(P(\text{B}) = 0.143; P(\text{N}) = 0.85\)
2. \(P(\text{N|B}) = 0.02\)
3. \(P(\text{B AND N}) = P(\text{B})P(\text{N|B}) = (0.143)(0.02) = 0.0029\)
4. \(P(\text{B OR N}) = P(\text{B}) + P(\text{N}) - P(\text{B AND N}) = 0.143 + 0.85 - 0.0029 = 0.9901\)
5. No. \(P(\text{N}) = 0.85; P(\text{N|B}) = 0.02\). So, \(P(\text{N|B})\) does not equal \(P(\text{N})\).
6. No. \(P(\text{B AND N}) = 0.0029\). For \(\text{B}\) and \(\text{N}\) to be mutually exclusive, \(P(\text{B AND N})\) must be zero

Exercise\(\PageIndex{5}\)

Refer to the information in Example \(\PageIndex{3}\). \(\text{P} =\) tests positive.

1. Given that a woman develops breast cancer, what is the probability that she tests positive. Find \(P(\text{P|B}) = 1 - P(\text{N|B})\).
2. What is the probability that a woman develops breast cancer and tests positive. Find \(P(\text{B AND P}) = P(\text{P|B})P(\text{B})\).
3. What is the probability that a woman does not develop breast cancer. Find \(P(\text{B′}) = 1 - P(\text{B})\).
4. What is the probability that a woman tests positive for breast cancer. Find \(P(\text{P}) = 1 - P(\text{N})\).

Answer

a. 0.98; b. 0.1401; c. 0.857; d. 0.15

Exercise \(\PageIndex{6}\)

A school has 200 seniors of whom 140 will be going to college next year. Forty will be going directly to work. The remainder are taking a gap year. Fifty of the seniors going to college play sports. Thirty of the seniors going directly to work play sports. Five of the seniors taking a gap year play sports. What is the probability that a senior is going to college and plays sports?

Answer

Let \(\text{A} =\) student is a senior going to college.

Let \(\text{B} =\) student plays sports.

\(P(\text{B}) = \dfrac{140}{200}\)

\(P(\text{B|A}) = \dfrac{50}{140}\)

\(P(\text{A AND B}) = P(\text{B|A})P(\text{A})\)

\(P(\text{A AND B}) = (\dfrac{140}{200}\))(\(\dfrac{50}{140}) = \dfrac{1}{4}\)

Contingency Tables (also called Two-way Tables)

A contingency table provides a way of portraying data that can facilitate calculating probabilities.  As we saw in the previous section, these tables help in determining probability of compound events, especially conditional probabilities, quite easily. The table displays sample values in relation to two different variables that may be dependent or contingent on one another. Later on, we will use contingency tables again, but in another manner.

Example \(\PageIndex{4}\)

Suppose a study of speeding violations and drivers who use cell phones produced the following fictional data:

	Speeding violation in the last year	No speeding violation in the last year	Total
Cell phone user	25	280	305
Not a cell phone user	45	405	450
Total	70	685	755

The total number of people in the sample is 755. The row totals are 305 and 450. The column totals are 70 and 685. Notice that 305 + 450 = 755 and 70 + 685 = 755.

Calculate the following probabilities using the table.

1. Find \(P(\text{Person is a cell phone user})\).
2. Find \(P(\text{person had no violation in the last year})\).
3. Find \(P(\text{Person had no violation in the last year AND was a cell phone user})\).
4. Find \(P(\text{Person is a cell phone user OR person had no violation in the last year})\).
5. Find \(P(\text{Person is a cell phone user GIVEN person had a violation in the last year})\).
6. Find \(P(\text{Person had no violation last year GIVEN person was not a cell phone user})\)

Solutions

1. \(\dfrac{\text{number of cell phone users}}{\text{total number in study}}\) = \(\dfrac{305}{755}\)
2. \(\dfrac{\text{number that had no violation}}{\text{total number in study}} = \dfrac{685}{755}\)
3. \(\dfrac{280}{755}\)
4. \(\left(\dfrac{305}{755} + \dfrac{685}{755}\right) - \dfrac{280}{755} = \dfrac{710}{755}\)
5. \(\dfrac{25}{70}\) (The sample space is reduced to the number of persons who had a violation.)
6. \(\dfrac{405}{450}\) (The sample space is reduced to the number of persons who were not cell phone users.)

Exercise \(\PageIndex{7}\)

Table shows the number of athletes who stretch before exercising and how many had injuries within the past year.

	Injury in last year	No injury in last year	Total
Stretches	55	295	350
Does not stretch	231	219	450
Total	286	514	800

1. What is \(P(\text{athlete stretches before exercising})\)?
2. What is \(P(\text{athlete stretches before exercising|no injury in the last year})\)?

Answer

1. \(P(\text{athlete stretches before exercising}) = \dfrac{350}{800} = 0.4375\)
2. \(P(\text{athlete stretches before exercising|no injury in the last year}) = \dfrac{295}{514} = 0.5739\)

Example \(\PageIndex{5}\)

Table shows a random sample of 100 hikers and the areas of hiking they prefer.

Hiking Area Preference

Sex	The Coastline	Near Lakes and Streams	On Mountain Peaks	Total
Female	18	16	___	45
Male	___	___	14	55
Total	___	41	___	___

1. Complete the table.
2. Are the events "being female" and "preferring the coastline" independent events? Let F = being female and let C = preferring the coastline.
   1. Find P(F AND C).
   2. Find P(F)P(C)
   3. Are these two numbers the same? If they are, then F and C are independent. If they are not, then F and C are not independent.
3. Find the probability that a person is male given that the person prefers hiking near lakes and streams. Let \(\text{M} =\) being male, and let \(\text{L} =\) prefers hiking near lakes and streams.
   1. What word tells you this is a conditional?
   2. Fill in the blanks and calculate the probability: \(P\)(___|___) = ___.
   3. Is the sample space for this problem all 100 hikers? If not, what is it?
4. Find the probability that a person is female or prefers hiking on mountain peaks. Let \(\text{F} =\) being female, and let \(\text{P} =\) prefers mountain peaks.
   1. Find \(P(\text{F})\).
   2. Find \(P(\text{P})\).
   3. Find \(P(\text{F AND P})\).
   4. Find \(P(\text{F OR P})\).

Solutions

a.

Hiking Area Preference

Sex	The Coastline	Near Lakes and Streams	On Mountain Peaks	Total
Female	18	16	11	45
Male	16	25	14	55
Total	34	41	25	100

b.

\(P(\text{F AND C}) = \dfrac{18}{100} = 0.18\)

\(P(\text{F})P(\text{C}) = \left(\dfrac{45}{100}\right) \left(\dfrac{34}{100}\right) = (0.45)(0.34) = 0.153\)

\(P(\text{F AND C}) \neq P(\text{F})P(\text{C})\), so the events \(\text{F}\) and \(\text{C}\) are not independent.

c.

1. The word 'given' tells you that this is a conditional.
2. \(P(\text{M|L}) = \dfrac{25}{41}\)
3. No, the sample space for this problem is the 41 hikers who prefer lakes and streams.

d.

1. Find \(P(\text{F})\).
2. Find \(P(\text{P})\).
3. Find \(P(\text{F AND P})\).
4. Find \(P(\text{F OR P})\).

d.

1. \(P(\text{F}) = \dfrac{45}{100}\)
2. \(P(\text{P}) = \dfrac{25}{100}\)
3. \(P(\text{F AND P}) = \dfrac{11}{100}\)
4. \(P(\text{F OR P}) = \dfrac{45}{100} + \dfrac{25}{100} - \dfrac{11}{100} = \dfrac{59}{100}\)

Exercise \(\PageIndex{8}\)

Table shows a random sample of 200 cyclists and the routes they prefer. Let \(\text{M} =\) males and \(\text{H} =\) hilly path.

Gender	Lake Path	Hilly Path	Wooded Path	Total
Female	45	38	27	110
Male	26	52	12	90
Total	71	90	39	200

1. Out of the males, what is the probability that the cyclist prefers a hilly path?
2. Are the events “being male” and “preferring the hilly path” independent events?

Answer

1. P(H|M) = \(\dfrac{52}{90}\) = 0.5778
2. For M and H to be independent, show P(H|M) = P(H)
   P(H|M) = 0.5778, P(H) = \(\dfrac{90}{200}\) = 0.45
   P(H|M) does not equal P(H) so M and H are NOT independent.

Example \(\PageIndex{6}\)

Muddy Mouse lives in a cage with three doors. If Muddy goes out the first door, the probability that he gets caught by Alissa the cat is \(\dfrac{1}{5}\) and the probability he is not caught is \(\dfrac{4}{5}\). If he goes out the second door, the probability he gets caught by Alissa is \(\dfrac{1}{4}\) and the probability he is not caught is \(\dfrac{3}{4}\). The probability that Alissa catches Muddy coming out of the third door is \(\dfrac{1}{2}\) and the probability she does not catch Muddy is \(\dfrac{1}{2}\). It is equally likely that Muddy will choose any of the three doors so the probability of choosing each door is \(\dfrac{1}{3}\).

Door Choice

Caught or Not	Door One	Door Two	Door Three	Total
Caught	\(\dfrac{1}{15}\)	\(\dfrac{1}{12}\)	\(\dfrac{1}{6}\)	____
Not Caught	\(\dfrac{4}{15}\)	\(\dfrac{3}{12}\)	\(\dfrac{1}{6}\)	____
Total	____	____	____	1

• The first entry \(\dfrac{1}{15} = \left(\dfrac{1}{5}\right) \left(\dfrac{1}{3}\right)\) is \(P(\text{Door One AND Caught})\)
• The entry \(\dfrac{4}{15} = \left(\dfrac{4}{5}\right) \left(\dfrac{1}{3}\right)\) is \(P(\text{Door One AND Not Caught})\)

Verify the remaining entries.

1. Complete the probability contingency table. Calculate the entries for the totals. Verify that the lower-right corner entry is 1.
2. What is the probability that Alissa does not catch Muddy?
3. What is the probability that Muddy chooses Door One OR Door Two given that Muddy is caught by Alissa?

Solution

Door Choice

Caught or Not	Door One	Door Two	Door Three	Total
Caught	\(\dfrac{1}{15}\)	\(\dfrac{1}{12}\)	\(\dfrac{1}{6}\)	\(\dfrac{19}{60}\)
Not Caught	\(\dfrac{4}{15}\)	\(\dfrac{3}{12}\)	\(\dfrac{1}{6}\)	\(\dfrac{41}{60}\)
Total	\(\dfrac{5}{15}\)	\(\dfrac{4}{12}\)	\(\dfrac{2}{6}\)	1

b. \(\dfrac{41}{60}\)

c. \(\dfrac{9}{19}\)

Exercise \(\PageIndex{9}\)

Table contains the number of crimes per 100,000 inhabitants from 2008 to 2011 in the U.S.

United States Crime Index Rates Per 100,000 Inhabitants 2008–2011

Year	Robbery	Burglary	Rape	Vehicle	Total
2008	145.7	732.1	29.7	314.7
2009	133.1	717.7	29.1	259.2
2010	119.3	701	27.7	239.1
2011	113.7	702.2	26.8	229.6
Total

TOTAL each column and each row. Total data = 4,520.7

1. Find \(P(\text{2009 AND Robbery})\).
2. Find \(P(\text{2010 AND Burglary})\).
3. Find \(P(\text{2010 OR Burglary})\).
4. Find \(P(\text{2011|Rape})\).
5. Find \(P(\text{Vehicle|2008})\).

Answer

a. 0.0294, b. 0.1551, c. 0.7165, d. 0.2365, e. 0.2575

Exercise \(\PageIndex{10}\)

Table relates the weights and heights of a group of individuals participating in an observational study.

Weight/Height	Tall	Medium	Short	Totals
Obese	18	28	14
Normal	20	51	28
Underweight	12	25	9
Totals

1. Find the total for each row and column
2. Find the probability that a randomly chosen individual from this group is Tall.
3. Find the probability that a randomly chosen individual from this group is Obese and Tall.
4. Find the probability that a randomly chosen individual from this group is Tall given that the idividual is Obese.
5. Find the probability that a randomly chosen individual from this group is Obese given that the individual is Tall.
6. Find the probability a randomly chosen individual from this group is Tall and Underweight.
7. Are the events Obese and Tall independent?

Answer

Weight/Height	Tall	Medium	Short	Totals
Obese	18	28	14	60
Normal	20	51	28	99
Underweight	12	25	9	46
Totals	50	104	51	205

1. Row Totals: 60, 99, 46. Column totals: 50, 104, 51.
2. \(P(\text{Tall}) = \dfrac{50}{205} = 0.244\)
3. \(P(\text{Obese AND Tall}) = \dfrac{18}{205} = 0.088\)
4. \(P(\text{Tall|Obese}) = \dfrac{18}{60} = 0.3\)
5. \(P(\text{Obese|Tall}) = \dfrac{18}{50} = 0.36\)
6. \(P(\text{Tall AND Underweight}) = \dfrac{12}{205} = 0.0585\)
7. No. \(P(\text{Tall})\) does not equal \(P(\text{Tall|Obese})\).

Probability of Multiple Events

So far, we have focused on the probability of a singular even occurring. It might have been compound, in which case we used the Addition Rule, or it might have been conditional, in which case we used the Multiplication Rule. However, we haven't yet used the Multiplication Rule to calculate the probability a two or more events. For these types of probabilities we need a generalized definition of Independent Events to help us identify when adjustments to our calculations need to be made without adding more complexity to our formulas. 

As a reminder, the condition for independence could be satisfied in a few ways. 

1. The events do not affect each other. 

Examples of this include guessing on answers for a multiple choice test, rolling a fair die, flipping a coin, responses from unassociated people, etc. 

2. Possible outcomes are replaced after a selection is made. 

Examples of this include picking an object out of a container and putting it back before the next selection, allowing people to be selected only once, etc. 

3. The sample size is 5% or less of the total population, then selections can be treated as independent events since the removal of an outcome doesn't overly affect the final calculation. 

An example of this could be randomly sampling 200 people and asking the question, what is the probability that 8 of them completed high school. 5% of 200 is 10. Since 8 is less than 10, we could treat our calculations as independent even though we are not replacing the individuals.  

Example \(\PageIndex{7}\)

You randomly guess the answer to four multiple-choice test questions (each with five possible choices). 

a. What is the probability of getting all four questions correct?

b. What is the probability of getting the first three questions correct and the last question wrong?

c. What is the probability of getting at least one question correct?

Solution

Let's define our event \(\text{E}\) to be getting the question right. So, \(P(\text{E}) = \frac{1}{5}\) since ther are 5 possible choices and only 1 is correct. 

a. Getting four questions correct means we did the following: E and E and E and E.

The "and" is the same one we use for our multiplication rule. We want to know what's the probability of getting the fourth question right given that we got the third question right given that we got the second question right given that we got the first question right. As you can probably imagine, expanding the formula to show that process makes it look a little complex.

However, since we already know that the probability of guessing the second question's answer isn't affected by our first guess, and so on, so these events are independent, which means that \(\text{P(E|E}) = P(\text{E})\). So our multiplication formula will look can be written simply as 

\[P(\text{E AND E AND E AND E}) = P(\text{E}) \cdot P(\text{E}) \cdot P(\text{E}) \cdot P(\text{E}) \nonumber\]

\[P(\text{E AND E AND E AND E}) = \frac{1}{5} \cdot \frac{1}{5} \cdot \frac{1}{5} \cdot \frac{1}{5} \nonumber\]

\[P(\text{E AND E AND E AND E}) = (\dfrac{1}{5})^4  \nonumber\]

\[P(\text{E AND E AND E AND E}) = \frac{1}{625} = 0.0016  \nonumber\]

There is a 0.16% chance of guessing all four questions correctly. 

 

b. Getting the first three questions correct and the last one wrong, otherwise known as not correct: E and E and E and E'. 

The complement of E, E', has four possible outcomes. Since there are four incorrect choices of the five. So, P\(\text{E}) = \frac{4}{5}\). We could also use our complement rule: \(P(\text{E}) + P(\text{E'}) = 1 \)

\[P(\text{E AND E AND E AND E}) = P(\text{E}) \cdot P(\text{E}) \cdot P(\text{E}) \cdot P(\text{E'}) \nonumber\]

\[P(\text{E AND E AND E AND E}) = \frac{1}{5} \cdot \frac{1}{5} \cdot \frac{1}{5} \cdot \frac{4}{5} \nonumber\]

\[P(\text{E AND E AND E AND E}) = (\dfrac{1}{5})^3 \cdot \dfrac{4}{5}  \nonumber\]

\[P(\text{E AND E AND E AND E}) = \frac{4}{625} = 0.0064  \nonumber\]

There is only a 0.64% chance of guessing three correct and then the last one wrong. 

 

c. Getting at least one question right means one or more answer is correct:

E and E and E and E where all of them are correct

E and E and E and E' where the last one is wrong, however, this outcome of 3 right and 1 wrong has three other variations

OR E and E and E' and E  OR E and E' and E and E' OR E and E and E and E. That makes this event a compound event, so we'll need to use the addition rule for each of the "OR" statements. But that means that in the case where two are correct and two are right, we will also have multiple outcomes. The same is true where one answer is correct and three are wrong. Instead of finding all of those additions and multiplications, let's think in terms of complements. 

The complement to "At least 1" is right is that none are right. \(P(\text{E} \geq 1) + P(\text{E} = 0) = 1 \) since this covers all possible outcomes for this event. So, \(P(\text{E} \geq 1) = 1 - P(\text{E} = 0)\) and  \(P(\text{E} = 0)\) just means we got none right so all of them were wrong. 

\[P(\text{E} = 0) = P(\text{E' AND E' AND E' AND E'}) = P(\text{E'}) \cdot P(\text{E'}) \cdot P(\text{E'}) \cdot P(\text{E'}) \nonumber\]

\[P(\text{E} = 0) =\frac{4}{5} \cdot \frac{4}{5} \cdot \frac{4}{5} \cdot \frac{4}{5} \nonumber\]

\[P(\text{E} = 0) =(\dfrac{4}{5})^4  =\frac{256}{625} = 0.4096 \nonumber\]

\[P(\text{E} \geq 1) = 1 - P(\text{E} = 0) \nonumber\]

\[P(\text{E} \geq 1) = 1 - 0.4096 = 0.5904 \nonumber\]

There is a 59.04% chance that a student will guess at least one question right on a four question multiple choice quiz. 

Exercise \(\PageIndex{11}\)

You roll two fair six-sided dice. What's the probability that the first number rolled is even and the second number rolled is a 5? 

Answer

The sample space is {1,2,3,4,5,6}. Let \(\text{A}\) be the even that the first number is even and  \(\text{B}\) be the event that a 5 is rolled. Since one die roll does not affect the next die roll, these are independent events.

\(P(\text{A and B} = P(\text{A}) \cdot P(\text{B}) = \frac{3}{6} \cdot \frac{1}{6} = \frac{3}{36}\)

For events where we are removing, or potentially removing, items from our sample space with each selection, we need to use an intuitive approach for our multiplication rule. Using it as demonstrated earlier requires some counting rules that we have not learned yet. However, doing so tends to overcomplicate the math so for our rule of 

\[P(\text{A AND B}) = P(B)P(A|B) \nonumber\]

we are going to amend the formula for P(A|B) to mean "what we assume happens next based on event B." 

Example \(\PageIndex{8}\)

A container holds three different colored marbles. Ten are blue, five are red and fifteen are green. There 30 total marbles.
a. If two marbles are selected with replacement, what is the probability that the first will be blue and the second will be blue?

b. If two marbles are selected without replacement, what is the probability that the first will be blue and the second will be blue?

Solution

a. First, we will continue to practice with independent events. Since we withdrew a marble and then put it back. The sample space for this experiment does not change when we reach in to select the second marble. Let's call \(\text{B} the event of drawing a blue marble.

\(P(\text{B and B}) = P(\text{B}) \cdot P(\text{B}) = \frac{10}{30} \cdot \frac{10}{30} = \frac{100}{900} = 0.1111 \) Rounded to 4 decimal places.

There is about an 11.11% chance of pulling a blue marble and then another blue marble. 

b. Now, we will no longer have independent events. Once we draw the first marble and don't put it back, then we only have 29 marbles remaining. Our sample space has changed and so does the probability of getting a blue marble. Additionally, if we assume that we already pulled a blue marble, then we only have 9 blue marbles let to pick from. In this case, we will use our \(P(\text{B|B}\) notation only to remind us of the change to our sample space. 

\(P(\text{B and B}) = P(\text{B}) \cdot P(\text{B|B}) = \frac{10}{30} \cdot \frac{9}{29} = \frac{90}{870} = 0.1034 \) Rounded to 4 decimal places.

There is about an 10.34% chance of pulling a blue marble and then another blue marble. As you might have expected, the probability of pulling another blue marble decreased since we had fewer to potentially pick from. 

Exercise \(\PageIndex{12}\)

You have a fair, well-shuffled deck of 52 cards. It consists of four suits. The suits are clubs, diamonds, hearts and spades. There are 13 cards in each suit consisting of 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, \(\text{J}\) (jack), \(\text{Q}\) (queen), \(\text{K}\) (king) of that suit.
a. If two cards are selected without replacement, what is the probability that the first will be a heart and the second will be a heart?

b. If two cards are selected without replacement, what is the probability that the first will be a heart and the second will be a spade?

Answer

a. Let \(\text{H}\) be the event of drawing a heart. Without replacement means that these events are dependent. \[P(\text{H and H}) = P(\text{H}) \cdot P(\text{H|H}) \nonumber\]  \[P(\text{H and H}) = \frac{13}{52} \cdot \frac{12}{51} = \frac{156}{2652} = 0.0588 \text{Rounded to 4 decimal places.} \nonumber\]

b. Let \(\text{S}\) be the event of drawing a spade. Without replacement means that these events are dependent. \[P(\text{H and S}) = P(\text{H})\cdot P(\text{S|H}) \nonumber\] \[P(\text{H and S}) = \frac{13}{52} \cdot \frac{13}{51} = \frac{169}{2652} = 0.0637 \text{Rounded to 4 decimal places.} \nonumber\]

Notice that \(P(\text{S|H})\) still had a numerator of 13 since the assumption was that we drew a heart the first time, so there should still be 13 spades in the deck.

Tree and Venn Diagrams

Sometimes, when the probability problems are complex, it can be helpful to graph the situation. Tree diagrams and Venn diagrams are two tools that can be used to visualize and solve conditional probabilities.

Tree Diagrams

A tree diagram is a special type of graph used to determine the outcomes of an experiment. It consists of "branches" that are labeled with either frequencies or probabilities. Tree diagrams can make some probability problems easier to visualize and solve. The following example illustrates how to use a tree diagram.

Example \(\PageIndex{9}\): Probabilities from Sampling with replacement

In an urn, there are 11 balls. Three balls are red (\(\text{R}\)) and eight balls are blue (\(\text{B}\)). Draw two balls, one at a time, with replacement (remember that "with replacement" means that you put the first ball back in the urn before you select the second ball). The tree diagram using frequencies that show all the possible outcomes follows.

Figure \(\PageIndex{1}\): Total = 64 + 24 + 24 + 9 = 121

The first set of branches represents the first draw. The second set of branches represents the second draw. Each of the outcomes is distinct. In fact, we can list each red ball as R1, R2, and R3 and each blue ball as B1, B2, B3, B4, B5, B6, B7, and B8. Then the nine RR outcomes can be written as:

R1R1 R1R2 R1R3 R2R1 R2R2 R2R3 R3R1 R3R2 R3R3

The other outcomes are similar.

There are a total of 11 balls in the urn. Draw two balls, one at a time, with replacement. There are 11(11) = 121 outcomes, the size of the sample space.

1. List the 24 BR outcomes: B1R1, B1R2, B1R3, ...
2. Using the tree diagram, calculate P(RR).
3. Using the tree diagram, calculate \(P(\text{RB OR BR})\).
4. Using the tree diagram, calculate \(P(\text{R on 1st draw AND B on 2nd draw})\).
5. Using the tree diagram, calculate P(R on 2nd draw GIVEN B on 1st draw).
6. Using the tree diagram, calculate \(P(\text{BB})\).
7. Using the tree diagram, calculate \(P(\text{B on the 2nd draw given R on the first draw})\).

Solution

1. B1R1; B1R2; B1R3; B2R1; B2R2; B2R3; B3R1; B3R2; B3R3; B4R1; B4R2; B4R3; B5R1; B5R2; B5R3; B6R1; B6R2; B6R3 B7R1; B7R2; B7R3; B8R1; B8R2; B8R3
2. \(P(\text{RR}) = \left(\frac{3}{11}\right) \left(\frac{3}{11}\right) = \frac{9}{121}\)
3. \(P(\text{RB OR BR}) = \left(\frac{3}{11}\right) \left(\frac{8}{11}\right) + \left(\frac{8}{11}\right) \left(\frac{3}{11}\right) = \frac{48}{121}\)
4. \(P(\text{R on 1st draw AND B on 2nd draw}) = P(\text{RB}) = \left(\frac{3}{11}\right) \left(\frac{8}{11}\right) = \frac{24}{121}\)
5. P(R on 2nd draw GIVEN B on 1st draw) = P(R on 2nd|B on 1st) = \(\frac{24}{88}\) = \(\frac{3}{11}\) This problem is a conditional one. The sample space has been reduced to those outcomes that already have a blue on the first draw. There are 24 + 64 = 88 possible outcomes (24 BR and 64 BB). Twenty-four of the 88 possible outcomes are BR. \(\frac{24}{88}\) = \(\frac{3}{11}\)
6. \(P(\text{BB}) = \frac{64}{121}\)
7. \(P(\text{B on 2nd draw|R on 1st draw}) = \frac{8}{11}\). There are 9 + 24 outcomes that have \(\text{R}\) on the first draw (9 RR and 24 RB). The sample space is then 9 + 24 = 33. 24 of the 33 outcomes have \(\text{B}\) on the second draw. The probability is then \(\frac{24}{33}\).

Exercise \(\PageIndex{13}\)

In a standard deck, there are 52 cards. 12 cards are face cards (event \(\text{F}\)) and 40 cards are not face cards (event \(\text{N}\)). Draw two cards, one at a time, with replacement. All possible outcomes are shown in the tree diagram as frequencies. Using the tree diagram, calculate \(P(\text{FF})\).

Figure \(\PageIndex{2}\):

Answer

Total number of outcomes is 144 + 480 + 480 + 1600 = 2,704.

\[P(\text{FF}) = \frac{144}{144+480+480+1,600} = \frac{144}{2,704} = \frac{9}{169}\]

 

Example \(\PageIndex{10}\): Probabilities from Sampling without replacement

An urn has three red marbles and eight blue marbles in it. Draw two marbles, one at a time, this time without replacement, from the urn. (remember that "without replacement" means that you do not put the first ball back before you select the second marble). Following is a tree diagram for this situation. The branches are labeled with probabilities instead of frequencies. The numbers at the ends of the branches are calculated by multiplying the numbers on the two corresponding branches, for example, \(\left(\frac{3}{11}\right)\left(\frac{2}{10}\right) = \frac{6}{110}\).

Figure \(\PageIndex{3}\): Total \(= \frac{56+24+24+6}{110} = \frac{110}{110} = 1\)

If you draw a red on the first draw from the three red possibilities, there are two red marbles left to draw on the second draw. You do not put back or replace the first marble after you have drawn it. You draw without replacement, so that on the second draw there are ten marbles left in the urn.

Calculate the following probabilities using the tree diagram.

1. \(P(\text{RR}) =\) ________
2. Fill in the blanks: \(P(\text{RB OR BR}) = \left(\frac{3}{11}\right) \left(\frac{8}{10}\right) +\) (___)(___) \(= \frac{48}{110}\)
3. \(P(\text{R on 2nd|B on 1st}) =\)
4. Fill in the blanks: \(P(\text{R on 1st AND B on 2nd}) = P(\text{RB}) =\) (___)(___) \(= \frac{24}{110}\)
5. Find \(P(\text{BB})\).
6. Find \(P(\text{B on 2nd|R on 1st})\).

Answers

1. \(P(\text{RR}) = \left(\frac{3}{11}\right)\left(\frac{2}{10}\right) = \frac{6}{110}\)
2. \(P(\text{RB OR BR}) = \left(\frac{3}{11}\right)\left(\frac{8}{10}\right) + \left(\frac{8}{11}\right)\left(\frac{3}{10}\right) = \frac{48}{110}\)
3. \(P(\text{R on 2nd|B on 1st}) = \frac{3}{10}\)
4. \(P(\text{R on 1st AND B on 2nd}) = P(\text{RB}) = \left(\frac{3}{11}\right) \left(\frac{8}{10}\right) = \frac{24}{110}\)
5. \(P(\text{BB}) = \left(\frac{8}{11}\right)\left(\frac{7}{10}\right)\)
6. Using the tree diagram, \(P(\text{B on 2nd|R on 1st}) = P(\text{R|B}) = \frac{8}{10}\).

If we are using probabilities, we can label the tree in the following general way.

• \(P(\text{R|R})\) here means \(P(\text{R on 2nd|R on 1st})\)
• \(P(\text{B|R})\) here means \(P(\text{B on 2nd|R on 1st})\)
• \(P(\text{R|B})\) here means \(P(\text{R on 2nd|B on 1st})\)
• \(P(\text{B|B})\) here means \(P(\text{B on 2nd|B on 1st})\)

Exercise \(\PageIndex{14}\)

In a standard deck, there are 52 cards. Twelve cards are face cards (\(\text{F}\)) and 40 cards are not face cards (\(\text{N}\)). Draw two cards, one at a time, without replacement. The tree diagram is labeled with all possible probabilities.

Figure \(\PageIndex{4}\):

1. Find \(P(\text{FN OR NF})\).
2. Find \(P(\text{N|F})\).
3. Find \(P(\text{at most one face card})\).
   Hint: "At most one face card" means zero or one face card.
4. Find \(P(\text{at least on face card})\).
   Hint: "At least one face card" means one or two face cards.

Answer

1. \(P(\text{FN OR NF}) = \frac{480}{2,652} + \frac{480}{2,652} = \frac{960}{2,652} = \frac{80}{221}\)
2. \(P(\text{N|F}) = \frac{40}{51}\)
3. \(P(\text{at most one face card}) = \frac{(480+480+1,560)}{2,652} = \frac{2,520}{2,652}\)
4. \(P(\text{at least one face card}) = \frac{(132+480+480)}{2,652} = \frac{1,092}{2,652}\)

Example \(\PageIndex{11}\)

A litter of kittens available for adoption at the Humane Society has four tabby kittens and five black kittens. A family comes in and randomly selects two kittens (without replacement) for adoption.

1. What is the probability that both kittens are tabby?
   a. \(\left(\frac{1}{2}\right) \left(\frac{1}{2}\right)\) b. \(\left(\frac{4}{9}\right) \left(\frac{4}{9}\right)\) c. \(\left(\frac{4}{9}\right) \left(\frac{3}{8}\right)\) d. \(\left(\frac{4}{9}\right) \left(\frac{5}{9}\right)\)
2. What is the probability that one kitten of each coloring is selected?
   a. \(\left(\frac{4}{9}\right) \left(\frac{5}{9}\right)\) b. \(\left(\frac{4}{9}\right) \left(\frac{5}{8}\right)\) c. \(\left(\frac{4}{9}\right) \left(\frac{5}{9}\right)\) + \(\left(\frac{5}{9}\right) \left(\frac{4}{9}\right)\) d. \(\left(\frac{4}{9}\right) \left(\frac{5}{8}\right)\) + \(\left(\frac{5}{9}\right) \left(\frac{4}{8}\right)\)
3. What is the probability that a tabby is chosen as the second kitten when a black kitten was chosen as the first?
4. What is the probability of choosing two kittens of the same color?

Answer

a. c, b. d, c. \(\frac{4}{8}\), d. \(\frac{32}{72}\)

Exercise \(\PageIndex{15}\)

Suppose there are four red balls and three yellow balls in a box. Three balls are drawn from the box without replacement. What is the probability that one ball of each coloring is selected?

Answer

\(\left(\frac{4}{7}\right) \left(\frac{3}{6}\right)\) + \(\left(\frac{3}{7}\right) \left(\frac{4}{6}\right)\)

Venn Diagram

A Venn diagram is a picture that represents the outcomes of an experiment. It generally consists of a box that represents the sample space S together with circles or ovals. The circles or ovals represent events.

Example \(\PageIndex{12}\)

Suppose an experiment has the outcomes 1, 2, 3, ... , 12 where each outcome has an equal chance of occurring. Let event \(\text{A} =\) {1, 2, 3, 4, 5, 6} and event \(\text{B} =\) {6, 7, 8, 9}. Then \(\text{A AND B} =\) {6} and \(\text{A OR B} =\) {1, 2, 3, 4, 5, 6, 7, 8, 9}. The Venn diagram is as follows:

Figure \(\PageIndex{5}\):

Exercise \(\PageIndex{16}\)

Suppose an experiment has outcomes black, white, red, orange, yellow, green, blue, and purple, where each outcome has an equal chance of occurring. Let event \(\text{C} =\) {green, blue, purple} and event \(\text{P} =\) {red, yellow, blue}. Then \(\text{C AND P} =\) {blue} and \(\text{C OR P} =\) {green, blue, purple, red, yellow}. Draw a Venn diagram representing this situation.

Answer

Figure \(\PageIndex{6}\):

Example \(\PageIndex{13}\)

Flip two fair coins. Let \(\text{A} =\) tails on the first coin. Let \(\text{B} =\) tails on the second coin. Then \(\text{A} =\) {TT,TH} and \(\text{B} =\) {TT, HT}. Therefore, \(\text{A AND B} =\) {TT}. \(\text{A OR B} =\) {TH, TT, HT}.

The sample space when you flip two fair coins is \(X =\) {HH, HT, TH, TT}. The outcome HH is in \(\text{NEITHER A NOR B}\). The Venn diagram is as follows:

Figure \(\PageIndex{7}\):

Exercise \(\PageIndex{17}\)

Roll a fair, six-sided die. Let \(\text{A} =\) a prime number of dots is rolled. Let \(\text{B} =\) an odd number of dots is rolled. Then \(\text{A} =\) {2, 3, 5} and \(\text{B} =\) {1, 3, 5}. Therefore, \(\text{A AND B} =\) {3, 5}. \(\text{A OR B} =\) {1, 2, 3, 5}. The sample space for rolling a fair die is \(\text{S} =\) {1, 2, 3, 4, 5, 6}. Draw a Venn diagram representing this situation.

Answer

Example \(\PageIndex{14}\): Probability and Venn Diagrams

Forty percent of the students at a local college belong to a club and 50% work part time. Five percent of the students work part time and belong to a club. Draw a Venn diagram showing the relationships. Let \(\text{C} =\) student belongs to a club and \(\text{PT} =\) student works part time.

Figure \(\PageIndex{8}\):

If a student is selected at random, find

• the probability that the student belongs to a club. \(P(\text{C}) = 0.40\)
• the probability that the student works part time. \(P(\text{PT}) = 0.50\)
• the probability that the student belongs to a club AND works part time. \(P(\text{C AND PT}) = 0.05\)
• the probability that the student belongs to a club given that the student works part time. \(P(\text{C|PT}) = \frac{P(\text{C AND PT})}{P(\text{PT})} = \frac{0.05}{0.50} = 0.1\)
• the probability that the student belongs to a club OR works part time. \(P(\text{C OR PT}) = P(\text{C}) + P(\text{PT}) - P(\text{C AND PT}) = 0.40 + 0.50 - 0.05 = 0.85\)

Exercise \(\PageIndex{18}\)

Fifty percent of the workers at a factory work a second job, 25% have a spouse who also works, 5% work a second job and have a spouse who also works. Draw a Venn diagram showing the relationships. Let \(\text{W} =\) works a second job and \(\text{S} =\) spouse also works.

Answer

Figure \(\PageIndex{9}\):

Example \(\PageIndex{15}\)

A person with type O blood and a negative Rh factor (Rh-) can donate blood to any person with any blood type. Four percent of African Americans have type O blood and a negative RH factor, 5−10% of African Americans have the Rh- factor, and 51% have type O blood.

Figure \(\PageIndex{10}\):

The “O” circle represents the African Americans with type O blood. The “Rh-“ oval represents the African Americans with the Rh- factor.

We will take the average of 5% and 10% and use 7.5% as the percent of African Americans who have the Rh- factor. Let \(\text{O} =\) African American with Type O blood and \(\text{R} =\) African American with Rh- factor.

1. \(P(\text{O}) =\) ___________
2. \(P(\text{R}) =\) ___________
3. \(P(\text{O AND R}) =\) ___________
4. \(P(\text{O OR R}) =\) ____________
5. In the Venn Diagram, describe the overlapping area using a complete sentence.
6. In the Venn Diagram, describe the area in the rectangle but outside both the circle and the oval using a complete sentence.

Answer

a. 0.51; b. 0.075; c. 0.04; d. 0.545; e. The area represents the African Americans that have type O blood and the Rh- factor. f. The area represents the African Americans that have neither type O blood nor the Rh- factor.

Exercise \(\PageIndex{19}\)

In a bookstore, the probability that the customer buys a novel is 0.6, and the probability that the customer buys a non-fiction book is 0.4. Suppose that the probability that the customer buys both is 0.2.

1. Draw a Venn diagram representing the situation.
2. Find the probability that the customer buys either a novel or anon-fiction book.
3. In the Venn diagram, describe the overlapping area using a complete sentence.
4. Suppose that some customers buy only compact disks. Draw an oval in your Venn diagram representing this event.

Answer

a. and d. In the following Venn diagram below, the blue oval represent customers buying a novel, the red oval represents customer buying non-fiction, and the yellow oval customer who buy compact disks.

Figure \(\PageIndex{11}\):

b. \(P(\text{novel or non-fiction}) = P(\text{Blue OR Red}) = P(\text{Blue}) + P(\text{Red}) - P(\text{Blue AND Red}) = 0.6 + 0.4 - 0.2 = 0.8\).

c. The overlapping area of the blue oval and red oval represents the customers buying both a novel and a nonfiction book.

References

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4. “Roulette.” Wikipedia. Available online at http://en.Wikipedia.org/wiki/Roulette (accessed May 2, 2013).
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6. Data from the Baseball-Almanac, 2013. Available online at www.baseball-almanac.com (accessed May 2, 2013).
7. Data from U.S. Census Bureau.
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Review

The multiplication rule and the addition rule are used for computing the probability of \(\text{A}\) and \(\text{B}\), as well as the probability of \(\text{A}\) or \(\text{B}\) for two given events \(\text{A}\), \(\text{B}\) defined on the sample space. In sampling with replacement each member of a population is replaced after it is picked, so that member has the possibility of being chosen more than once, and the events are considered to be independent. In sampling without replacement, each member of a population may be chosen only once, and the events are considered to be not independent. The events \(\text{A}\) and \(\text{B}\) are mutually exclusive events when they do not have any outcomes in common.

A tree diagram use branches to show the different outcomes of experiments and makes complex probability questions easy to visualize. A Venn diagram is a picture that represents the outcomes of an experiment. It generally consists of a box that represents the sample space S together with circles or ovals. The circles or ovals represent events. A Venn diagram is especially helpful for visualizing the OR event, the AND event, and the complement of an event and for understanding conditional probabilities.

Formula Review

The multiplication rule: \(P(\text{A AND B}) = P(\text{A|B})P(\text{B})\)

The addition rule: \(P(\text{A OR B}) = P(\text{A}) + P(\text{B}) - P(\text{A AND B})\)

Glossary

Independent Events

The occurrence of one event has no effect on the probability of the occurrence of another event. Events \(\text{A}\) and \(\text{B}\) are independent if one of the following is true:

1. \(P(\text{A|B}) = P(\text{A})\)
2. \(P(\text{B|A}) = P(\text{B})\)
3. \(P(\text{A AND B}) = P(\text{A})P(\text{B})\)

Mutually Exclusive

Two events are mutually exclusive if the probability that they both happen at the same time is zero. If events \(\text{A}\) and \(\text{B}\) are mutually exclusive, then \(P(\text{A AND B}) = 0\).

Contingency table (also called Two-way Table)

the method of displaying a frequency distribution as a table with rows and columns to show how two variables may be dependent (contingent) upon each other; the table provides an easy way to calculate conditional probabilities.

Tree Diagram

the useful visual representation of a sample space and events in the form of a “tree” with branches marked by possible outcomes together with associated probabilities (frequencies, relative frequencies)

Venn Diagram

the visual representation of a sample space and events in the form of circles or ovals showing their intersections