We are given a basket containing 12 apples, of which 3 are rotten, and 9 are good. We are asked to find the probability of drawing at most one rotten apple when 3 apples are drawn at random.
Step 1:
The total number of ways to select 3 apples out of 12 is given by the combination formula:
\[
\binom{12}{3} = \frac{12 \times 11 \times 10}{3 \times 2 \times 1} = 220
\]
Step 2:
We need to find the probability of drawing at most one rotten apple, which means there can be either 0 or 1 rotten apple in the selection of 3 apples.
Case 1: Drawing 0 rotten apples (all 3 apples are good):
The number of ways to select 3 good apples from 9 is:
\[
\binom{9}{3} = \frac{9 \times 8 \times 7}{3 \times 2 \times 1} = 84
\]
Case 2: Drawing 1 rotten apple (2 good apples and 1 rotten apple):
The number of ways to select 1 rotten apple from 3 and 2 good apples from 9 is:
\[
\binom{3}{1} \times \binom{9}{2} = 3 \times \frac{9 \times 8}{2 \times 1} = 3 \times 36 = 108
\]
Step 3:
Now, the total number of favorable outcomes is the sum of the two cases:
\[
84 + 108 = 192
\]
Step 4:
The probability of drawing at most one rotten apple is the ratio of favorable outcomes to total outcomes:
\[
P(\text{at most one rotten apple}) = \frac{192}{220} = \frac{48}{55}
\]
Thus, the probability that at most one rotten apple is drawn is \( \frac{48}{55} \).