Question

Exercise 2 A box contains 3 white balls, 4 red balls and 5 black balls. A ball is picked, its color recorded and returned to the box(with replacement). Another ball is then selected and its color recorded.

1. Find the probability that 2 black balls are selected.

2. Find the probability that 2 balls of the same color are selected.

Now 4 balls are picked with replacement

3.Find the probability no red balls are selected.

4.Find the probability that the fourth ball selected is the first occurrence of the color white?

Answer 1

Number of white balls = 3

Number of red balls = 4

Number of black balls = 5

Total number of balls = 3 + 4 + 5 = 12

1. P(2 black balls are selected) = (5/12)²

= 0.1736

2. P(2 balls of same color) = (3/12)² + (4/12)² + (5/12)²

= 0.3472

3. P(red ball) = 4/12 = 1/3

P(not a red ball) = 1 - 1/3 = 2/3

P(no red balls) = (2/3)⁴

= 0.1975

P(white) = 3/12 = 0.25

P(not white) = 1 - 0.25 = 0.75

P(fourth ball selected is the first occurrence of the color white) = P(first 3 balls are not white) x P(fourth ball is white)

= 0.75³ x 0.25

= 0.1055