Three Persons A, B And C Shoot To Hit A Target. If In Trials, A Hits The Target 4 Times In 5 Shots, B Hits 3 Times In 4 Shots And C Hits 2 Times In 3 Trials. Find The Probability That Exactly Two Persons Hit The Target

To solve this problem, we first need to understand the individual probabilities of each person hitting the target.

1. Person A hits the target 4 times out of 5. So, the probability of A hitting the target, P(A) = 4/5 = 0.8 and not hitting the target, P(A') = 1 - P(A) = 0.2.

2. Person B hits the target 3 times out of 4. So, the probability of B hitting the target, P(B) = 3/4 = 0.75 and not hitting the target, P(B') = 1 - P(B) = 0.25.

3. Person C hits the target 2 times out of 3. So, the probability of C hitting the target, P(C) = 2/3 = 0.67 and not hitting the target, P(C') = 1 - P(C) = 0.33.

We are asked to find the probability that exactly two persons hit the target. This can happen in three ways:

1. A and B hit the target but C does not.
2. A and C hit the target but B does not.
3. B and C hit the target but A does not.

We calculate the probability for each scenario and then add them up.

1. The probability that A and B hit the target but C does not is P(A) * P(B) * P(C') = 0.8 * 0.75 * 0.33 = 0.198.

2. The probability that A and C hit the target but B does not is P(A) * P(C) * P(B') = 0.8 * 0.67 * 0.25 = 0.134.

3. The probability that B and C hit the target but A does not is P(B) * P(C) * P(A') = 0.75 * 0.67 * 0.2 = 0.1.

Adding these probabilities together gives us the total probability that exactly two persons hit the target, which is 0.198 + 0.134 + 0.1 = 0.432. So, the probability that exactly two persons hit the target is 0.432.