The probability that A hits a target is 1/4 and the probability that B hits a target 1/3. (a) If each fires twice, what is the probability that the target will be hit at least once? (b) If each fire once and the target is hit only once, what is the probability that A hit the target? Solution probability that A hits a target = 1/4 probability that B hits a target = 1/3 a) after hitting twice, probability that the target will be hit at least once = 1 - probability that the target is not hit at all probability that the target is not hit at all = Probability that A does not hit the target the first time * Probability that B does not hit the target the first time * Probability that A does not hit the target the second time * Probability that B does not hit the target the second time = (1 - 1/4) * (1 - 1/3) * (1 - 1/4) * (1 - 1/3) = 3/4 * 2/3 * 3/4 * 2/3 = 1/4 after hitting twice, probability that the target will be hit at least once = 1 - 1/4 = 3/4 ------------------------------------------ b)If each fire once and the target is hit only once, the probability that A hit the target = probability that A hit the target * probability that B missed + probability that A hit the target * probability that B hit the target = 1/4 * (1 - 1/3) + 1/4 * 1/3 = 1/4 * 2/3 + 1/4 * 1/3 = 1/4.

1. The probability that A hits a target is 1/4 and the probability that B hits a target 1/3.
(a) If each fires twice, what is the probability that the target will be hit at least once?
(b) If each fire once and the target is hit only once, what is the probability that A hit the target?
Solution
probability that A hits a target = 1/4
probability that B hits a target = 1/3
a) after hitting twice, probability that the target will be hit at least once = 1 - probability that the
target is not hit at all
probability that the target is not hit at all = Probability that A does not hit the target the first time
* Probability that B does not hit the target the first time * Probability that A does not hit the
target the second time * Probability that B does not hit the target the second time
= (1 - 1/4) * (1 - 1/3) * (1 - 1/4) * (1 - 1/3)
= 3/4 * 2/3 * 3/4 * 2/3
= 1/4
after hitting twice, probability that the target will be hit at least once = 1 - 1/4 = 3/4
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b)If each fire once and the target is hit only once, the probability that A hit the target =
probability that A hit the target * probability that B missed + probability that A hit the target *
probability that B hit the target
= 1/4 * (1 - 1/3) + 1/4 * 1/3
= 1/4 * 2/3 + 1/4 * 1/3
= 1/4