The probabilities that three men hit a target are 1/6, 1/4, 1/3respectively. Each shoots once thetarget. What is the probability that exactly oneof them hits the target?a)23/96b)42/127c)19/48d)31/72Correct answer is option 'D'. Can you explain this answer?

Question

Answer

Problem:
The probabilities that three men hit a target are 1/6, 1/4, 1/3 respectively. Each shoots once at the target. What is the probability that exactly one of them hits the target?

Solution:

To solve this problem, we need to find the probability that exactly one of the three men hits the target.

Let's denote the events as follows:
- A: The first man hits the target
- B: The second man hits the target
- C: The third man hits the target

We are given the probabilities of each man hitting the target as follows:
- P(A) = 1/6
- P(B) = 1/4
- P(C) = 1/3

We need to find the probability of exactly one of these events occurring. To do this, we can use the following formula:

P(Exactly one event occurs) = P(A) * (1 - P(B)) * (1 - P(C)) + (1 - P(A)) * P(B) * (1 - P(C)) + (1 - P(A)) * (1 - P(B)) * P(C)

Let's substitute the given probabilities into the formula:

P(Exactly one event occurs) = (1/6) * (1 - 1/4) * (1 - 1/3) + (1 - 1/6) * (1/4) * (1 - 1/3) + (1 - 1/6) * (1 - 1/4) * (1/3)

Simplifying the equation:

P(Exactly one event occurs) = (1/6) * (3/4) * (2/3) + (5/6) * (1/4) * (2/3) + (5/6) * (3/4) * (1/3)

P(Exactly one event occurs) = 1/4 + 5/24 + 5/24

P(Exactly one event occurs) = 6/24 + 5/24 + 5/24

P(Exactly one event occurs) = 16/24

Simplifying the fraction:

P(Exactly one event occurs) = 2/3

Therefore, the probability that exactly one of the three men hits the target is 2/3, which is not one of the given answer choices.

Hence, the given answer is incorrect.

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