Practice Multiplication Rule in Statistics

Use these practice problems to test your method after reviewing the concept explanation and worked examples.

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Quick Recap

The multiplication rule finds the probability that two events both occur. It multiplies the probability of the first event by the conditional probability of the second event given that the first has happened.

For an ā€œandā€ problem, move through the events in sequence. Take the chance of the first step, then update for the second step based on what is already known.

Showing a random 20 of 50 problems.

Example 1

medium
Five cards are dealt. Find P(firstĀ isĀ ace,Ā secondĀ isĀ king)P(\text{first is ace, second is king}).

Example 2

easy
For independent A,BA,B with P(A)=0.3P(A)=0.3, P(B)=0.6P(B)=0.6, find P(Aāˆ©B)P(A\cap B).

Example 3

hard
From 55 couples (1010 people), pick 22 people at random. Find P(theyĀ areĀ aĀ couple)P(\text{they are a couple}).

Example 4

easy
Find P(Aāˆ©B)P(A\cap B) if P(A)=0.5P(A)=0.5 and P(Bāˆ£A)=0.4P(B\mid A)=0.4.

Example 5

medium
A test has P(disease)=0.01P(\text{disease})=0.01 and P(positiveāˆ£disease)=0.99P(\text{positive}\mid\text{disease})=0.99. Find P(diseaseĀ andĀ positive)P(\text{disease and positive}).

Example 6

easy
P(A)=0.4P(A) = 0.4 and P(B)=0.5P(B) = 0.5 are independent. Find P(Aāˆ©B)P(A\cap B).

Example 7

challenge
From 5 red and 5 blue balls, draw 2 without replacement. Find P(atĀ leastĀ oneĀ red)P(\text{at least one red}).

Example 8

medium
A student passes math with probability 0.70.7. If they pass math, they pass physics with probability 0.80.8; otherwise with probability 0.40.4. Find P(passesĀ both)P(\text{passes both}).

Example 9

challenge
In a class of 2323 people, find P(atĀ leastĀ 2Ā shareĀ aĀ birthday)P(\text{at least }2\text{ share a birthday}), assuming 365365 equally likely birthdays.

Example 10

medium
A bag has 2 red, 3 blue, 5 green (10 total). Draw 2 without replacement. Find P(redĀ thenĀ green)P(\text{red then green}).

Example 11

easy
P(A)=0.5P(A) = 0.5, P(Bāˆ£A)=0.6P(B\mid A) = 0.6. Find P(Aāˆ©B)P(A\cap B).

Example 12

medium
From a deck, 33 cards are dealt without replacement. Find P(allĀ 3Ā areĀ spades)P(\text{all }3\text{ are spades}).

Example 13

easy
A coin and a die: find P(tailsĀ andĀ primeĀ number)P(\text{tails and prime number}) assuming independence (primes among 11ā€“66 are 2,3,52,3,5).

Example 14

easy
P(A)=0.3P(A) = 0.3 and P(Bāˆ£A)=0.7P(B\mid A) = 0.7. Find P(Aāˆ©B)P(A\cap B).

Example 15

medium
A drawer has 6 black and 4 white socks. Draw 2 without replacement. Find P(bothĀ sameĀ color)P(\text{both same color}).

Example 16

medium
A factory line: part passes inspection 1 with probability 0.90.9, and if it passes 1 it passes inspection 2 with probability 0.80.8. Find P(passesĀ both)P(\text{passes both}).

Example 17

easy
A bag has 44 red and 66 blue. Draw 22 without replacement. Find P(bothĀ red)P(\text{both red}).

Example 18

medium
P(Aāˆ©B)=0.12P(A\cap B)=0.12 and P(A)=0.4P(A)=0.4. Find P(Bāˆ£A)P(B\mid A).

Example 19

easy
From a deck, draw 22 cards without replacement. Find P(bothĀ kings)P(\text{both kings}).

Example 20

easy
A box has 5 good and 1 defective fuse. Draw 2 without replacement. Find P(bothĀ good)P(\text{both good}).
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