Question
Download Solution PDFLet A and B be two players who are playing the game to hit the target. The probabilities of hitting the target by A and B is \(\frac{2}{3}\) and \(\frac{3}{4}\), respectively. What is the probability that exactly one of them hit the target?
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFGiven:
Probability of A hitting the target, P(A) = 2/3
Probability of B hitting the target, P(B) = 3/4
Formula used:
Probability of A not hitting the target, P(A') = 1 - P(A)
Probability of B not hitting the target, P(B') = 1 - P(B)
Probability of exactly one hitting = P(A)×P(B') + P(A')×P(B)
Calculations:
P(A') = 1 - 2/3 = 1/3
P(B') = 1 - 3/4 = 1/4
Probability of exactly one hitting = (2/3)×(1/4) + (1/3)×(3/4)
⇒ 2/12 + 3/12 = 5/12
∴ The probability that exactly one of them hits the target is 5/12.
Last updated on Jul 8, 2025
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