4.4: Multiplication Rules and Conditional Probability
- Page ID
- 45475
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- 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
When two events A and B are independent, then \( P(A \text{ and } B) = P(A) \cdot P(B)\)
A coin is tossed once, followed by the rolling of a single die. What is the probability of getting a head on the coin and a 5 on the die?
Sample space for the Problem ={ (Heads, 1), (Heads, 2), (Heads, 3), (Heads, 4), (Heads, 5), (Heads, 6), (Tails, 1), (Tails, 2), (Tails, 3), (Tails, 4), (Tails, 5), (Tails, 6)}
Solution
There are a total of 12 possible outcomes, and 1 of them is heads and 5. Thus, the probability is P(heads and 5) = \(\dfrac{1}{12}\).
Using the formula above, we get the same answer. is P(heads and 5) = \(\dfrac{1}{2} \cdot \dfrac{1}{6} = \dfrac{1}{12} \).
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.
- \(P (\text{ First is a 7 and Second is a 7} ) = \dfrac{6}{36} \cdot \dfrac{6}{36} = \dfrac{1}{6} \cdot \dfrac{1}{6} = \dfrac{1}{36} = 0.028 \)
- \(P (\text{ First is a 5 and Second is an 8} ) = \dfrac{4}{36} \cdot \dfrac{5}{36} = \dfrac{1}{9} \cdot \dfrac{5}{36} = \dfrac{5}{324} = 0.015 \)
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.
\(P (\text{ All three pass the class} ) = 0.48 \cdot 0.48 \cdot 0.48 = 0.48^3 = 0.111\)
A local elementary school is hiring new staff, and each applicant has a 72% chance of being hired. Four Citrus College students majoring in child development apply for positions at the school. Assuming each hiring decision is independent, what is the probability that all four students are hired?
Solution
\(P (\text{ All four are hired at the elementary school} ) = 0.72 \cdot 0.72\cdot 0.72\cdot 0.72= 0.72^4 = 0.269\)
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.
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.
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.
\(P (\text{ All 3 Cards are Jacks} ) = \dfrac{4}{52} \cdot \dfrac{3}{51}\cdot \dfrac{2}{50} = \dfrac{1}{5525} = 0.000181\)
Three scholarships of $1000 are to be awarded to three liberal arts majors. The liberal arts majors who qualify for the scholarship consist of 7 economics majors, 4 sociology majors, 3 political science majors, and 6 anthropology majors. Compute the chance that all three scholarships will be awarded to the sociology 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.
\(P (\text{ All 3 are sociology Majors} ) = \dfrac{4}{20} \cdot \dfrac{3}{19}\cdot \dfrac{2}{18} = \dfrac{1}{285} = 0.00351\)
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.
The probability that event B will occur given that A has occurred can be computed as follows.
\(P(B|A) = \dfrac{P(A \text{ and } B)}{P(A)}\)
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 \(P( \text{ Second is a king| First is a king}) = \dfrac{3}{51} = 0.059\)
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.
\(P(\text{First is Red}) = \dfrac{5}{12}\)
\(P(\text{First and Second is Red}) = \dfrac{5}{12} \cdot \dfrac{4}{11} = \dfrac{20}{132}\)
\( P(\text{Second is Red | First is Red}) = \dfrac{P(\text{First and Second are Red})}{P(\text{First is Red})} = \dfrac{\dfrac{20}{132}}{\dfrac{5}{12}} = \dfrac{4}{11} = 0.364 \)
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.
| A | B | AB | O | Total | |
|---|---|---|---|---|---|
| Male | 61 | 81 | 30 | 96 | 268 |
| Female | 24 | 70 | 92 | 84 | 270 |
| Total | 85 | 151 | 122 | 180 | 538 |
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. \(P(\text {Female|Blood types B or AB}) = \dfrac{162}{273} = 0.593\)
Exercises
Two fair six-sided dice are rolled.
- Event A: The first die shows a 4
- Event B: The second die shows a 6
Since the dice are rolled independently, the events are independent.
What is the probability that the first die shows a 4 and the second die shows a 6?
- Answer
-
Since these events are independent, use the multiplication rule I.
P(first die is a 4 and second die is a 6) = \( \dfrac{1}{6}\cdot\dfrac{1}{6} = \dfrac{1}{36}\)
An economist is studying consumer behavior.
- The probability that a randomly selected consumer prefers online shopping is 0.60
- The probability that a randomly selected consumer uses digital payment methods is 0.70
Assume these two events are independent.
What is the probability that a randomly selected consumer prefers online shopping and uses digital payment methods?
- Answer
-
Since these events are independent, use the multiplication rule I.
P(prefers online shopping and uses digital payment methods) = \( 0.6 \cdot 0.7 = 0.42\)
An economist is analyzing income and investment behavior.
- The probability that a randomly selected individual is in the top income bracket is 0.20
- The probability that a randomly selected individual owns stocks is 0.50
What is the probability that a randomly selected individual is in the top income bracket and owns stocks?
- Answer
-
Since these events are independent, use the multiplication rule I.
P(top income bracket and owns stocks) = \( 0.2 \cdot 0.5 = 0.10\)
A standard deck of 52 playing cards is used. Two cards are drawn without replacement.
- Event A: The first card drawn is a Heart
- Event B: The second card drawn is also a Heart
What is the probability that both cards drawn are Hearts?
- Answer
-
Since these events are dependent, use the multiplication rule II.
P(both cards drawn are Hearts) = \( \dfrac{13}{52}\cdot\dfrac{12}{51} = \dfrac{1}{17}\)
A university is awarding scholarships to students from different majors.
- 10 are child development majors
- 7 are economics majors
- 5 are sociology majors
Three students are selected at random without replacement.
- What is the probability that the first student selected is a child development major, the second is an economics major, and the third is a sociology major?
- What is the probability that all 3 are economics majors?
- Answer
-
Since these events are dependent, use the multiplication rule II.
- P (child development major, economics major, and sociology major) = \( \dfrac{10}{22}\cdot\dfrac{7}{21}\cdot\dfrac{5}{20} = \dfrac{5}{132}\)
- P (all 3 are economics majors) = \( \dfrac{7}{22}\cdot\dfrac{6}{21}\cdot\dfrac{5}{20} = \dfrac{1}{44}\)
A standard deck of 52 playing cards is used. One card is randomly selected.
- Event A: The card is a Face Card (Jack, Queen, or King)
- Event B: The card is a Heart
What is the probability that the card is a Face Card given that it is a Heart?
- Answer
-
The condition is that the first card is a heart, and there are 13 hearts. Of those 13 hearts, 3 of them are face cards.
Thus, the answer is P ( face card | heart) = \(\dfrac{3}{13}\)
A sociologist surveys 200 adults about community involvement and education level.
The results are shown in the table below:
| College Degree | No College Degree | Total | |
| Volunteer | 60 | 40 | 100 |
| Does Not Volunteer | 30 | 70 | 100 |
| Total | 90 | 110 | 200 |
What is the probability that a person volunteers, given that they have a college degree?
- Answer
-
Focus on the column for a college degree. Of those 90, 60 volunteer.
P (volunteer | college degree) = \(\dfrac{60}{90} = \dfrac{2}{3}\)
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