We are given a box of 24 apples, of which 6 are bad. We are taking three apples 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

We are given a box of 24 apples, of which 6 are bad. We are taking three apples 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 determine the number of good apples:

$24 - 6 = 18$.

Then we calculate the probability of picking a good apple:

$P(good) = \frac{18}{24} = \frac{3}{4}$

And the probability of picking a bad apple:

$P(bad) = \frac{6}{24} = \frac{1}{4}$

(a) The first two are good and the third is bad. Since we are drawing with replacement, the draws are independent.

$P(good, good, bad) = P(good) * P(good) * P(bad) = \frac{3}{4} * \frac{3}{4} * \frac{1}{4} = \frac{9}{64}$

(b) All three are bad.

$P(bad, bad, bad) = P(bad) * P(bad) * P(bad) = \frac{1}{4} * \frac{1}{4} * \frac{1}{4} = \frac{1}{64}$

(c) All three are good.

$P(good, good, good) = P(good) * P(good) * P(good) = \frac{3}{4} * \frac{3}{4} * \frac{3}{4} = \frac{27}{64}$

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 $\frac{27}{64}$.