Multiplication Rule Examples in Statistics

Start with the recap, study the fully worked examples, then use the practice problems to check your understanding of Multiplication Rule.

This page combines explanation, solved examples, and follow-up practice so you can move from recognition to confident problem-solving in Statistics.

Concept 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 “and” 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.

Read the full concept explanation →

How to Use These Examples

Read the first worked example with the solution open so the structure is clear.
Try the practice problems before revealing each solution.
Use the related concepts and background knowledge badges if you feel stuck.

What to Focus On

Core idea: Multiplication Rule starts by naming the possible outcomes and the event rule before assigning or combining probabilities.

Common stuck point: Students often know a procedure related to multiplication rule but skip the recognition step: Am I reasoning about what can happen and how likely it is, with the correct sample space or condition? That leads to a calculation or graph that looks reasonable but answers a different question.

Sense of Study hint: Ask: Am I reasoning about what can happen and how likely it is, with the correct sample space or condition?

Common Mistakes to Watch For

Before you work through the examples, skim the mistake guide so you know which shortcuts and sign errors to avoid.

Conditional Probability Mistakes

Worked Examples

Example 1

easy

A box has

7

good and

3

broken bulbs. Two are drawn without replacement. Find

P(\text{both good})

Answer

\dfrac{7}{15}

First step

P(1\text{st good}) = 7/10

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Example 2

medium

A class has

12

girls and

8

boys. Two students are chosen at random without replacement. Find

P(\text{both girls})

Class: 12 girls, 8 boys — choose 2 students without replacement

Example 3

medium

From a deck,

3

cards are dealt without replacement. Find

P(\text{all }3\text{ are spades})

Example 4

medium

A factory produces

5\%

defective items. Three items are selected independently. Find

P(\text{none defective})

Example 5

medium

A student passes math with probability

0.7

. If they pass math, they pass physics with probability

0.8

; otherwise with probability

0.4

. Find

P(\text{passes both})

Student: P(pass math)=0.7; P(pass physics | pass math)=0.8

Example 6

hard

A jar has

4

red,

5

blue,

6

green. Draw

3

without replacement. Find

P(\text{one of each color, in any order})

Example 7

hard

A bag has

3

red,

4

blue,

5

green. Draw

2

without replacement. Find

P(\text{same color})

Example 8

hard

In a

3

-stage interview process, an applicant passes each stage with probability

0.8

given they passed the previous. Find

P(\text{passes all }3)

3-stage interview: P(pass each stage | passed previous) = 0.8

Example 9

hard

A coin has

P(H) = 0.6

. Three flips. Find

P(\text{HHT in that order})

Example 10

challenge

10

people randomly take

10

named coats. Find

P(\text{no one gets their own coat})

for small case... actually use the derangement count for

n=4

to find

P(\text{no fixed points among }4\text{ people})

Practice Problems

Try these problems on your own first, then open the solution to compare your method.

Example 1

easy

Find

P(A\cap B)

P(A)=0.5

and

P(B\mid A)=0.4

Example 2

easy

A bag has 3 red and 2 blue balls. Draw 2 without replacement. Find

P(\text{both red})

Bag: 3 red, 2 blue — draw 2 without replacement

Example 3

easy

Draw 2 cards without replacement. Find

P(\text{both aces})

Example 4

easy

P(A)=0.6

, and given

A

P(B\mid A)=0.5

. Find

P(A\text{ and }B)

Example 5

easy

A jar has 4 green and 6 yellow candies. Draw 2 without replacement. Find

P(\text{green then yellow})

Jar: 4 green, 6 yellow — draw 2 without replacement

Example 6

easy

For independent

A,B

with

P(A)=0.3

P(B)=0.6

, find

P(A\cap B)

Example 7

easy

A box has 5 good and 1 defective fuse. Draw 2 without replacement. Find

P(\text{both good})

Box: 5 good, 1 defective — draw 2 without replacement

Example 8

easy

A coin and a die: find

P(\text{heads and even})

treating them as independent.

Example 9

medium

Draw 3 cards without replacement. Find

P(\text{all hearts})

Example 10

medium

A factory line: part passes inspection 1 with probability

0.9

, and if it passes 1 it passes inspection 2 with probability

0.8

. Find

P(\text{passes both})

Example 11

medium

A bag has 2 red, 3 blue, 5 green (10 total). Draw 2 without replacement. Find

P(\text{red then green})

Bag: 2 red, 3 blue, 5 green — draw 2 without replacement

Example 12

medium

A test has

P(\text{disease})=0.01

and

P(\text{positive}\mid\text{disease})=0.99

. Find

P(\text{disease and positive})

Example 13

medium

Two cards drawn without replacement. Find

P(\text{first king, second queen})

Example 14

medium

A spinner lands on win with probability

0.3

. To win a prize you must win on two independent spins. Find

P(\text{prize})

Example 15

medium

A drawer has 6 black and 4 white socks. Draw 2 without replacement. Find

P(\text{both same color})

Example 16

medium

P(A\cap B)=0.12

and

P(A)=0.4

. Find

P(B\mid A)

Example 17

medium

A coin is flipped and a card is drawn (independent). Find

P(\text{heads and a heart})

Example 18

challenge

A box has 7 good and 3 bad bulbs. Draw 3 without replacement. Find

P(\text{all good})

Example 19

challenge

From 5 red and 5 blue balls, draw 2 without replacement. Find

P(\text{at least one red})

Example 20

challenge

A class is

60\%

girls. Among girls

30\%

play sports; among boys

50\%

play. Find

P(\text{girl and plays sports})

and

P(\text{plays sports})

Class: 60% girls, 30% of girls play sports; 40% boys, 50% of boys play sports

Example 21

easy

P(A) = 0.3

and

P(B\mid A) = 0.7

. Find

P(A\cap B)

Example 22

easy

A coin and a die: find

P(\text{tails and prime number})

assuming independence (primes among

1

–

6

are

2,3,5

Example 23

easy

From a deck, draw

2

cards without replacement. Find

P(\text{both kings})

Example 24

easy

A spinner shows red with probability

0.4

. Spun twice. Find

P(\text{red then not red})

assuming independence.

Example 25

easy

A bag has

3

green and

5

yellow candies. Draw

2

without replacement. Find

P(\text{yellow then green})

Bag: 3 green, 5 yellow — draw 2 without replacement

Example 26

medium

A drug works on

80\%

of patients with the disease. Of those who improve,

90\%

recover fully. Find

P(\text{works AND fully recovers})

Drug: works with P=0.8; given works, full recovery P=0.9

Example 27

medium

A bag has

6

white and

4

black. Draw

3

without replacement. Find

P(\text{all white})

Example 28

medium

P(A) = 0.4

P(B\mid A) = 0.5

P(C\mid A\cap B) = 0.6

. Find

P(A\cap B\cap C)

Example 29

medium

A bag has

5

red,

4

blue,

1

green (

10

total). Draw

2

without replacement. Find

P(\text{red then blue})

Example 30

medium

In a class of

30

(

18

girls,

12

boys), pick

2

different students at random. Find

P(\text{first girl, second boy})

Class: 18 girls, 12 boys — pick 2 students without replacement

Example 31

hard

From a deck,

4

cards are dealt without replacement. Find

P(\text{all }4\text{ are aces})

Example 32

hard

Four dice are rolled. Find

P(\text{all four show the same number})

Example 33

hard

From

5

couples (

10

people), pick

2

people at random. Find

P(\text{they are a couple})

Example 34

challenge

In a class of

23

people, find

P(\text{at least }2\text{ share a birthday})

, assuming

365

equally likely birthdays.

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Related Concepts

Conditional Probability Tree Diagram Independent Events Expected Value

Background Knowledge

These ideas may be useful before you work through the harder examples.

conditional probabilitytree diagram