Question

A basket contains 12 apples in which 3 are rotten. If 3 apples are drawn at random simultaneously from it, then the probability of getting at most one rotten apple is:

• A. $\frac{34}{55}$
• B. $\frac{48}{55}$
• C. $\frac{21}{55}$
• D. $\frac{42}{55}$

Hint:

When calculating probabilities involving combinations, first find the total number of outcomes, then calculate the favorable outcomes for each case and sum them. Finally, divide the favorable outcomes by the total outcomes to find the probability.

Answer 1

The Correct Option is B

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 $\frac{48}{55}$.