Question

A can hit a target 4 times in 5 shots, B hits 3 times in 4 shots, and C hits twice in 3 shots. They fire together. Find the probability that at least two shots hit the target.

• A. \( \frac{13}{30} \)
• B. \( \frac{5}{6} \)
• C. \( \frac{11}{40} \)
• D. None of these

Hint:

Always simplify the probabilities before performing the calculations to avoid errors in complex operations.

Answer 1

The Correct Option is B

We are asked to find the probability that at least two shots hit the target when A, B, and C fire together. This can be calculated by first finding the total probability for all possible outcomes and then subtracting the probability where fewer than two shots hit the target. The individual probabilities for A, B, and C hitting the target are: - A: \( \frac{4}{5} \) - B: \( \frac{3}{4} \) - C: \( \frac{2}{3} \) We can calculate the probability that fewer than two shots hit the target (i.e., 0 or 1 hit), and then subtract it from 1 to get the probability of at least two hits. The total probability of at least two hits is calculated as follows: \[ P(ABC') + P(A'BC) + P(AB'C) + P(ABC) \] Now, calculating each individual term: \[ P(ABC') = \frac{4}{5} \times \frac{3}{4} \times \frac{1}{3} = \frac{1}{5} \] \[ P(A'BC) = \frac{1}{5} \times \frac{3}{4} \times \frac{2}{3} = \frac{2}{15} \] \[ P(AB'C) = \frac{4}{5} \times \frac{1}{4} \times \frac{2}{3} = \frac{2}{15} \] \[ P(ABC) = \frac{4}{5} \times \frac{3}{4} \times \frac{2}{3} = \frac{2}{5} \] Thus, adding the probabilities: \[ P = \frac{1}{5} + \frac{2}{15} + \frac{2}{15} + \frac{2}{5} = \frac{5}{6} \] Thus, the probability that at least two shots hit the target is \( \frac{5}{6} \).