There are 24 apples in a box, 6 of which are bad. Three apples are taken from the box at random, with replacement. We need to find the probability of the following events: (a) The first two apples are good and the third is bad. (b) All three apples are bad. (c) All three apples are good.

1. Problem Description

There are 24 apples in a box, 6 of which are bad. Three apples are taken from the box at random, with replacement. We need to find the probability of the following events:

(a) The first two apples are good and the third is bad.

(b) All three apples are bad.

(c) All three apples are good.

2. Solution Steps

First, we need to find the probability of picking a good apple and a bad apple.

Total number of apples = 24

Number of bad apples = 6

Number of good apples = 24 - 6 = 18

Probability of picking a good apple, $P(G) = \frac{18}{24} = \frac{3}{4}$

Probability of picking a bad apple, $P(B) = \frac{6}{24} = \frac{1}{4}$

(a) The first two are good and the third is bad. Since we are replacing the apples, the draws are independent. The probability is:

$P(GGB) = P(G) \times P(G) \times P(B) = \frac{3}{4} \times \frac{3}{4} \times \frac{1}{4} = \frac{9}{64}$

(b) All three are bad. Since we are replacing the apples, the draws are independent. The probability is:

$P(BBB) = P(B) \times P(B) \times P(B) = \frac{1}{4} \times \frac{1}{4} \times \frac{1}{4} = \frac{1}{64}$

(c) All three are good. Since we are replacing the apples, the draws are independent. The probability is:

$P(GGG) = P(G) \times P(G) \times P(G) = \frac{3}{4} \times \frac{3}{4} \times \frac{3}{4} = \frac{27}{64}$

3. Final Answer

(a) The probability that the first two are good and the third is bad is $\frac{9}{64}$.

(b) The probability that all three are bad is $\frac{1}{64}$.

(c) The probability that all three are good is $\frac{27}{64}$.