Complementary Events Examples - Word Problems

# Complementary Events Examples

Complementary events are those events who have possibility of two outcomes. If probability of getting one of the complementary event is p, then the probability of getting the other event will be,
q = (1 - p)

Example 1:

In tossing a coin, if the probability of getting a head is found to be 0.53, then what will be the probability of getting a tail?

Solution:

Getting a head or a tail are complementary events.

Hence, probability of getting a tail = (1 - 0.53) = 0.47

## Complementary Events Word Problems

Problem 1:

The probability of a bulb to be on is 0.34, then what will be the probability of that bulb to be switched off?

Solution:

Switching on and off of a bulb are complementary events.

The probability of the bulb to be switched off, $p = (1 - 0.34) = 0.66$
Problem 2:

Anita is playing the semi-final of chess tournament today, with her chances of no winning be 117117. Find her chances of winning.

Solution:

Winning or not winning a match are complementary events.

Probability of winning the match = $1$ - $\frac{1}{17}$ = $\frac{16}{17}$
Problem 3:

A bag contains only white and black balls. The probability of getting a white ball is 213213. Find the probability of getting a black ball if a ball is chosen at random.

Solution:

The bag can give only white or black ball. Hence, the event of getting a white ball is complementary to the event of getting a black ball.

Probability of getting a black ball = $1$ - probability of getting a white ball = $1$ - $\frac{2}{13}$ = $\frac{11}{13}$
Problem 4:

A die has been rolled. What is the probability of not getting a multiple of 3?

Solution:

Getting a multiple of 3 and not getting a multiple of 3 are complementary to each other.

Probability of getting a multiple of $3$ = $\frac{2}{6}$ = $\frac{1}{3}$

Probability of not getting a multiple of $3$ = $1$ - $\frac{1}{3}$ = $\frac{2}{3}$
Problem 5:

In a cricket match, a player hits a 4 in every 10 balls. Find the probability of him not hitting a 4.

Solution:

Probability of him hitting a $4$ = $\frac{4}{10}$ = $\frac{2}{5}$

As hitting a 4 and not hitting a 4 are complementary to each other,

Probability of him hitting a $4$ = $1$ - $\frac{2}{5}$ = $\frac{3}{5}$