Question

A box contains 10 balls, of which 3 are red, 2 are yellow, and 5 are blue. Five balls are randomly selected with replacement. Calculate the probability that fewer than 2 of the selected balls are red.

Answer 1

The required probability is 0.1.

Explanation:

red balls = 3

yellow balls = 2

blue balls = 5

Selected balls = 5

Number of elemnets in sample space = 10 C 5 = 1260

Ways to choose 1 red ball and 4 other colours = (3 C 1 ) x (7 C 4) = 105

Ways to choose 5 balls of other colours = 7 C 5 = 21

So, the probability is

`(105)/(1260) + \frac {21}{1260}\\\\(126)/(1260)=0.1`