# Multiplication Rule for Probabilities of Independent Events

\( \)\( \)\( \)\( \)Examples on using the multiplication rule to find the probability of two or more independent events occurring are presented along with detailed solutions.

## Independent Events

In probabilities, two events are independent if the occurrence of one does not affect the probability of occurrence of the other.

Example 1

The following events A and B independent.

- A = "roll a die and get a \( 1 \)" , B = "flip a coin and get a tail".
- A = "draw a card from a deck and get a King", replace it back into the deck, B = "draw another card and get a Queen"
- A = "roll a die and get a \( 4 \)" , B = "roll the same die (or another one) and get a "6"
- A = "flip a coin and get a head" , B = roll the same coin (or another one) and get a tail"

A jar has 3 blue balls, 2 white balls and 5 red balls - A = Pick a ball at random from the jar and get a red ball, replace it back into the jar, B = Pick a ball at random from the jar and get a white ball

Events C and D are NOT independent.

- C = "draw a card from a deck and get a King", D = "draw a second card from the same deck and get a Queen".

A jar contains 3 blue balls, 2 white balls and 3 red balls - C = "Pick a ball at random from the jar and get a red ball", D = "Pick a second ball at random from the same jar and get a white ball".

## Examples with Detailed Solutions

Example 2
A coin is tossed twice. What is the probability of getting a tail in the first toss and a tail in the second toss?

__Solution to Example 2__

Two methods to answer the question in example 2 are presented to show the advantage in using the product rule given above.

Method 1: __Using the sample space__

The sample space S of the experiment of tossing a coin twice is given by the tree diagram shown below

The first toss gives two possible outcomes : T or H ( in blue)

The second toss gives two possible outcomes : T or H (in red)

From the tree diagram, we can deduce the sample space \( S \) set as follows

\(
S = \{(H, H), (H, T), (T, H), (T,T) \}
\)

with \( n(S) = 4 \) where \( n(S) \) is the number of elements in the set \( S \)

The event \( E \) : " tossing a coin twice and getting two tails " as a set is given by

\(
E = \{(T,T) \}
\)

with \( n(E) = 1 \) where \( n(E) \) is the number of elements in the set \( E \)

Use the classical probability formula to find \( P(E) \) as:

\( P(E) = \dfrac{n(E)}{n(S)} = \dfrac{1}{4}\)

Method 2: Use the product rule of two independent event

Event \( E \) " tossing a coin twice and getting a tail in each toss " may be considered as two events

Event \( A \) " toss a coin once and get a tail " and event \( B \) "toss the coin a second time and get a tail "

with the probabilities of each event \( A \) and \(B \) given by

\( P(A) = \dfrac{1}{2} \) and \( P(B) = \dfrac{1}{2} \)

Event E occurring may now be considered as events A and B occurring. Events A and B are independent and therefore the product rule may be used as follows

\( P(E) = P( A \; and \; B) = P(A \cap B) = P(A) \cdot P(B) = \dfrac{1}{2} \cdot \dfrac{1}{2} = \dfrac{1}{4} \)

NOTE If you toss a coin a large number of times, the sample space will have a large number of elements and therefore method 2 is much more practical to use than method 1 where you have a large number of outcomes.

We now present more examples and questions on how the product rule of independent events is used to solve probability questions.

Example 3

A coin is tossed and a die is rolled. What is the probability of getting a head and a \( 4 \)?

__Solution to Example 3__

We have two independent events to consider:

Event A " toss a coin a get a head " and event B " roll a die and get a \( 4 \) "

When a coin is tossed, the probability of getting a head is

\( P(A) = \dfrac{1}{2} \)

When die is rolled, the probability of getting a \( 4 \) is

\( P(B) = \dfrac{1}{6} \)

\( P ( \) " getting a head and a \( 4 \) " \( ) = P( A \; and \; B) = P( A \cap B) = P(A) \cdot P(B) = \dfrac{1}{2} \cdot \dfrac{1}{6} = \dfrac{1}{12} \)

Example 4

A jar has 3 blue balls, 2 white balls and 5 red balls. A ball is selected at random and the color noted then replaced back inside the jar. A second ball is selected its color noted and replaced back inside the jar. A third ball is selected and its color noted.

What is the probability of

a) selecting 3 red balls

b) selecting a blue ball, then a white ball and then a blue ball

c) selecting a red, then a white ball and then a blue ball

__Solution to Example 4__

a)

Let event A "select a red ball the first time",

event B "select a red ball the second time"

and event C "select a red ball the third time"

All three events A, B and C are independent because the ball selected is replaced back into the jar.

Total number of balls is 10 and there are 5 red balls.

Let us now calculate the probability of selecting a red ball.

There are 5 red balls out of a total of 10, hence

\( P(A) \) = \( P(B) \) = \( P(C) \) =\( P(red) = \dfrac{5}{10} \\ = \dfrac{1}{2} \)

We use an extended formula to three independent events

\( P( \; A \; and \; B \; and \; C) = P(A) \cdot P(B) \cdot P(C) \\\\
= \dfrac{1}{2} \cdot \dfrac{1}{2} \cdot \dfrac{1}{2} = \dfrac{1}{8} \)

b)

Let event A "select a blue ball the first time",

event B "select a white ball the second time"

and event C "select a blue ball the third time"

\( P(A) = P(blue) = \dfrac{3}{10} \quad , \quad P(B) = P(white) = \dfrac{2}{10} = \dfrac{1}{5} \quad , \quad P(C) = P(blue) = \dfrac{3}{10}\)

\( P( A \; and \; B \; and \; C) = P(A) \cdot P(B) \cdot P(C) \\\\ = \dfrac{3}{10} \cdot \dfrac{1}{5} \cdot \dfrac{3}{10} \dfrac{3}{10} \\\\ = \dfrac{9}{5000} \)

c)

Let event A "select a red ball the first time",

event B "select a white ball the second time"

and event C "select a blue ball the third time"

\( P(A) = P(red) = 1/2 \)

\( P(B) = P(white) = 1/5 \)

\( P(C) = P(blue) = 3/10 \)

\( P( A \; and \; B \; and \; C) = P(A) \cdot P(B) \cdot P(C) \\\\ = \dfrac{1}{2} \cdot \dfrac{1}{5} \cdot \dfrac{3}{10} \\\\ = \dfrac{3}{100} \)

Example 5

A card is drawn from a deck of 52 cards and then replaced and a second card is drawn. Find the probability of getting a "2" and then a "King".

__Solution to Example 5__

We have two independent events to consider:

Event A " draw a card and get a 2" and event B " draw a card and get a King "

Because the card is replaced, the two events A and B are independent.

Let us first find \( P(A) \) and \( P(B) \).

\( P(A) = 4/52 = 1/13 \)

\( P(B) = 4/52 = 1/13 \)

\( P (A \; and \; B) = P(A) \cdot P(B) = \dfrac{1}{13} \cdot \dfrac{1}{13} = \dfrac{1}{169} \)

Example 6

A poll found that 25% of the people in a certain country have heart problems. If three people are selected at random, find the probability that all three have heart problems.

__Solution to Example 6__

Event A " the first person have heart problems" , event B " the second person has heart problems " and C " the third person has heart problems "

\( P(A) = 0.25\) , \( P(B) = 0.25 \) and \( P(C) = 0.25 \).

These are independent events, hence

\( P (A \; and \; B \; and \; C ) = P(A) \cdot P(B) \cdot P(C) = 0.25 \cdot 0.25 \cdot 0.25 = 0.015625 \)

### More Questions with Solutions

1) A die is rolled twice. Find the probability of getting an even number in the first throw and a number greater than 4 in the second throw.2) A card is drawn from a deck of 52 cards and then replaced and a second card is drawn. Find the probability of getting a "King" and then a "Queen of hearts".

3) In a country, 45% of the people smoke. If 4 people are selected at random, what is the population that they are all smokers?

### Solutions to above exercises

1)Let event A: get an even number, and event B: get a number greater than 4

A = \( \{2,4,6\} \) , B = \( \{5,6\} \)

\( P(A) = 3/6 = 1/2 \) , \( P(B) = 2/6 = 1/3 \)

Events A and B are independent; hence

\( P(A \; and \; B) = P(A) \cdot P(B) = \dfrac{1}{2} \cdot \dfrac{1}{3} = \dfrac{1}{6} \)

2)

\( P(King) = P(A) = 4/52 = 1/13 \) , \( P(Queen \; of \; hearts) = P(B) = 1/52 \)

Events A and B are independent; hence

\( P(A \; and \; B) = P(A) \cdot P(B) = \dfrac{1}{13} \cdot \dfrac{1}{52} = \dfrac{1}{676} \)

3)

\( P(\text{smoker}) = 0.45 \)

All 4 events are independent; hence

\( P(\text{all 4 smokers}) = P(\text{smoker} \; and \; \text{smoker} \; and \; \text{smoker} \; and \; \text{smoker}) \)

\( = P(\text{smoker}) \cdot P(\text{smoker}) \cdot P(\text{smoker}) \cdot P(\text{smoker}) \)

\( = (0.45)^4 = 0.04100625 \)

### More References and links

Addition Rule for Probabilities .Binomial Probabilities Examples and Questions

probability questions

classical formula for probability

mutually exclusive events

Introduction to Probabilities

sample space

event elementary statistics and probabilities .

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