4.4: Multiplication Rules and Conditional Probability

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Toros Berberyan, Tracy Nguyen, and Alfie Swan
Citrus College

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Learning Objectives

Use multiplication rules to determine the probability of two events occurring together.
Calculate the probability of independent events by multiplying their probabilities.
Apply conditional probability for dependent events, adjusting calculations based on prior occurrences.

In probability, independent events are events where the occurrence of one does not affect the occurrence of the other. For example, flipping a coin and rolling a die are independent events because the outcome of the coin flip (heads or tails) does not influence the number that appears on the die. Moreover, when selecting a small sample from a large population, individual selections can often be treated as independent events because the removal of one item does not significantly alter the probabilities of selecting subsequent items. This is particularly true when the sample size is much smaller than the total population.

Multiplication Rule#1

Definition: Multiplication Rule#1

When two events A and B are independent, then $ P(A \text{ and } B) = P(A) \cdot P(B)$

Example $\PageIndex{1}$

Assume that two dice are tossed twice. Compute the following probabilities. Write the answers as a fraction or a decimal rounded to three places.

Get a sum of 7 on both tosses.
Get a sum of 5 on the first toss and a sum of 8 on the second toss.

Solution

Since tossing two dice are independent events. The multiplication Rule#1 can be used to answer the question.

$P (\text{ First is a 7 and Second is a 7} ) = \dfrac{6}{36} \cdot \dfrac{6}{36} = \dfrac{1}{6} \cdot \dfrac{1}{6} = \dfrac{1}{36} = 0.028 $
$P (\text{ First is a 5 and Second is an 8} ) = \dfrac{4}{36} \cdot \dfrac{5}{36} = \dfrac{1}{9} \cdot \dfrac{5}{36} = \dfrac{5}{324} = 0.015 $

Example $\PageIndex{2}$

In the California Community College system, the chance of passing an introductory statistics course is around 48%. If three students are randomly selected from the system, what is the chance that all three will pass the class in one term? Write the answer as a decimal rounded to three places.

Solution

Since a small sample is selected from a large population, the events are independent, and Multiplication Rule#1 can be used to answer the question.

$P (\text{ All three pass the class} ) = 0.48 \cdot 0.48 \cdot 0.48 = 0.48^3 = 0.111$

Multiplication Rule#2

Dependent events are events where the outcome of one event affects the probability of the other occurring. This means that when one event happens, it changes the likelihood of the second event happening. For example, if you draw a card from a deck and do not put it back, the probability of drawing a second card of a certain type changes because the total number of cards in the deck has decreased.

Definition: Multiplication Rule#2

When two events A and B are dependent, the $ P(A \text{ and } B) = P(A) \cdot P(B|A)$

where $P(B|A)$ is the probability that B occurs given that A has occurred.

Example $\PageIndex{3}$

A person selects three cards from a standard deck without replacement. Compute the chance that all three cards are jacks. Write the answer as a decimal rounded to three non-zero decimal places.

Solution

Since the cards are selected without replacement, the events are dependent, and Multiplication Rule#2 is applied to solve the answer.

$P (\text{ All 3 Cards are Jacks} ) = \dfrac{4}{52} \cdot \dfrac{3}{51}\cdot \dfrac{2}{50} = \dfrac{1}{5525} = 0.000181$

Example $\PageIndex{4}$

Three scholarships of $500 are to be awarded to three STEM majors. The STEM majors who qualify for the scholarship consist of 7 physics majors, 4 math majors, 3 computer science majors, and 6 biology majors. Compute the chance that all three scholarships will be awarded to the math majors. Write the answer as a decimal rounded to three non-zero decimal places.

Solution

Since a person can only win the award once, the events are dependent, and Multiplication Rule#2 is applied to solve the answer.

$P (\text{ All 3 are Math Majors} ) = \dfrac{4}{20} \cdot \dfrac{3}{19}\cdot \dfrac{2}{18} = \dfrac{1}{285} = 0.00351$

Conditional Probabilities

Conditional probability is the probability of an event occurring given that another event has already happened. It helps us understand how the likelihood of one event changes based on new information. An example of dependent events is drawing two jacks from a standard deck of 52 cards without replacement. When you draw the first card, the probability of getting a jack is 4/52 since there are 4 jacks in the deck. If the first card is a jack, only 3 jacks remain, and the total number of cards is now 51. This means the probability of drawing a second jack is 3/51. Since the outcome of the first draw affects the probability of the second, these events are dependent.

Definition: Conditional Probability

The probability that event B will occur given that A has occurred can be computed as follows.

$P(B|A) = \dfrac{P(A \text{ and } B)}{P(A)}$

Example $\PageIndex{5}$ conditional probabilities

Suppose you select two cards from a standard deck without replacement. Find the following probabilities. Write the answer as a decimal rounded to three non-zero decimal places.

The second card is a queen, given that the first card is a jack.
The second card is a club, given that the first card is a heart.
The second card is a king, given that the first card is also a king.

Solutions

Since one jack has been removed from the deck, there is now a total of 51 cards and 4 queens. Thus the answer is $P( \text{ Second is a queen| First is a jack }) = \dfrac{4}{51} = 0.078$

Since one heart has been removed from the deck, there is now a total of 51 cards and 13 clubs. Thus the answer is $P( \text{ Second is a club| First is a heart }) = \dfrac{13}{51} = 0.255$
Since one king has been removed from the deck, there is now a total of 51 cards and 3 kings. Thus the answer is $P( \text{ Second is a king| First is a king}) = \dfrac{3}{51} = 0.059$

Example $\PageIndex{6}$

An urn contains 5 red marbles and 7 blue marbles, making a total of 12 marbles. Two marbles are drawn one after the other without replacement. What is the probability that the second marble drawn is red, given that the first marble drawn was red?

Solution

In this problem, the formula will be used to find the answer.

$P(\text{First is Red}) = \dfrac{5}{12}$

$P(\text{First and Second is Red}) = \dfrac{5}{12} \cdot \dfrac{4}{11} = \dfrac{20}{132}$

$ P(\text{Second is Red | First is Red}) = \dfrac{P(\text{First and Second are Red})}{P(\text{First is Red})} = \dfrac{\dfrac{20}{132}}{\dfrac{5}{12}} = \dfrac{4}{11} = 0.364 $

Example $\PageIndex{7}$

The contingency table below presents data from a survey conducted among students at a local college. The survey respondents are categorized by gender and blood type. Write the answer as a fraction or decimal rounded to 3 nonzero places.

Contingency Table for Blood Type and Gender
	A	B	AB	O	Total
Male	61	81	30	96	268
Female	24	70	92	84	270
Total	85	151	122	180	538

Table $\PageIndex{1}$: Contingency Table for Gender and Blood Type

If a student is selected, compute the following:

What is the probability that a student is a male given that the student has blood type B?
What is the probability that the student has blood type O, given that the student is a female?
What is the probability that the student is a female given that the student has blood type B or AB?

Solution

The condition is type B. Focus on the B column. $P(\text {Male|Blood type B}) = \dfrac{81}{151} = 0.536$
The condition is female. Focus on the female row. $P(\text {Blood type O|Female}) = \dfrac{84}{270} = 0.311$
The condition is blood type B or AB. Focus on the B or AB rows. $P(\text {Female|Blood types B or AB}) = \dfrac{162}{273} = 0.593$

Authors

"4.4: Multiplication Rules and Conditional Probability" by Toros Berberyan, Tracy Nguyen, and Alfie Swan is licensed under CC BY-SA 4.0

Attributions

"4.3: Conditional Probability" by Kathryn Kozak is licensed under CC BY-SA 4.0

Exercises

Determine whether the events in each example have a dependent or independent relationship. Explain your reasoning for each.
1. Owning a PlayStation 5 and owning an iPhone 15.
2. Passing your statistics class and studying for all quizzes and exams in the class.

Scan the QR code or click on it to open the MyOpenMath version of the above question with step-by-step guidance.
MyOpenMath is a free online learning platform designed to support math instruction through automated homework, quizzes, and assessments. You must register for MyOpenMath and sign in to view the question.

If two cards are drawn from a fair deck without replacement. Write the answer as a fraction or decimal rounded to 3 nonzero place values.
1. What is the probability that the second card is a Queen given that the first card is a Queen?
2. What is the probability that the second card is a heart given that the first card is a Spade?
3. What is the probability that both cards are Red cards?

At a popular restaurant, the probability that a person orders pizza and hot wings is 88%. The probability that the person will order only pizza is 98%. What is the probability that a person orders hot wings given that he or she orders pizza?

The probability that a student does all the homework and earns an A in the class is 85%. The probability that a student does all the homework is 90%. What is the probability that a student will earn an A given that he or she does all the homework? Write the answer as a percentage and round it to a whole percentage.

In the table below, the medal count for the Olympics is presented for five countries. Write the answer as a fraction or decimal rounded to 3 nonzero place values.

Contingency Table for Country and Medal Count
Country	Gold	Silver	Bronze	Total
USA	39	41	33	113
China	38	32	18	88
Japan	27	14	17	58
Germany	17	32	22	71
UK	22	21	22	65
Total	143	140	112	395

Table $\PageIndex{2}$: Contingency Table for Country and Medal Type

If an athlete or medal winner is selected, compute the following:

What is the probability that an athlete is from Germany given that he or she won a Silver Medal?
What is the probability that an athlete won a Bronze Medal given that he or she is from China?
What is the probability that an athlete won a Gold Medal given that he or she is from Japan or the UK?

The following contingency table lists 5 popular majors at a local community college according to gender. Write the answer as a fraction or decimal rounded to 3 nonzero place values.

Contingency Table for Gender and Major
Gender	Business	Computer Science	Engineering	Nursing	Psychology	Total
Female	94	94	41	43	22	294
Male	71	34	40	91	80	316
Total	165	128	81	134	102	610

Table $\PageIndex{3}$: Contingency Table for Gender and Major

If a student or a major is selected, compute the following:

What is the probability that the student is a male given that the student majored in engineering?
What is the probability that the student majored in nursing given that the student is a female?
What is the probability that the student is a female given that the student majored in business or computer science?

Answers

If you are an instructor and want the solutions to all the exercise questions for each section, please email Toros Berberyan.

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Definition: Multiplication Rule#1

Example \(\PageIndex{1}\)

Solution

Example \(\PageIndex{2}\)

Solution

Definition: Multiplication Rule#2

Example \(\PageIndex{3}\)

Solution

Example \(\PageIndex{4}\)

Solution

Definition: Conditional Probability

Example \(\PageIndex{5}\) conditional probabilities

Solutions

Example \(\PageIndex{6}\)

Solution

Example \(\PageIndex{7}\)

Solution