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:

The required probability is 0.1.

Step-by-step 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

Answer:The required probability is 0.1.

Step-by-step 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

[tex]\frac{105}{1260} + \frac {21}{1260}\\\\\frac{126}{1260}=0.1[/tex]