Al and Bill each have two bullets. They fire alternately at a glass bottle. Al has hitting probability 1/3 and Bill 1/4. What is the probability that Al smashes the bottle if he is the one who fires the first shot? (Note: Al can smash the bottle even if he misses with his first shot.)

Question

Talan Riggs

Mathematics

22 June, 07:00

Al and Bill each have two bullets. They fire alternately at a glass bottle. Al has hitting probability 1/3 and Bill 1/4. What is the probability that Al smashes the bottle if he is the one who fires the first shot? (Note: Al can smash the bottle even if he misses with his first shot.)

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Answers (1)

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Sawyer Melton · Answer 1

P (Al hits bottle first time) = 1/3

P (Al misses the first shot but hits on his second shot) =

P (Al misses and bill misses and Al hits) = 2/3 * 3/4 * 1/3 = 1/6

So required probability = 1/3 + 1/6 = 1/2