Let 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 2/3 and ¾, respectively. What is the probability that exactly one of them hit the target?
5/12
Given: \(P(A) = \dfrac{2}{3}\) and \(P(B) = \dfrac{3}{4}\).
The probability that exactly one of them hits the target is:
\(P(\text{exactly one}) = P(A)\cdot(1-P(B)) + (1-P(A))\cdot P(B)\)
Substituting the values:
\(= \dfrac{2}{3}\cdot\left(1-\dfrac{3}{4}\right) + \left(1-\dfrac{2}{3}\right)\cdot\dfrac{3}{4}\)
\(= \dfrac{2}{3}\cdot\dfrac{1}{4} + \dfrac{1}{3}\cdot\dfrac{3}{4}\)
\(= \dfrac{2}{12} + \dfrac{3}{12} = \dfrac{5}{12}\)
Hence, the required probability is 5/12.
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