On average, a sharpshooter hits the target once every 3 shots. What is the probability that he will hit
the target in 4 shots?
(A) 1
(B) 1/81
(C) 1/3
(D) 65/81
(E) 71/81
The sharpshooter hits the target once in 3 shots. Hence, the probability of hitting the target is 1/3. The
probability of not hitting the target is 1 – 1/3 = 2/3.
Now, (the probability of not hitting the target even once in 4 shots) + (the probability of hitting at least once
in 4 shots) equals 1, because these are the only possible cases.
Hence, the probability of hitting the target at least once in 4 shots is
1 – (the probability of not hitting even once in 4 shots)
The probability of not hitting in the 4 chances is
2/3.2/3.2/3.2/3=16/81 Now, 1 – 16/81 = 65/81. The answer is (D).
the target in 4 shots?
(A) 1
(B) 1/81
(C) 1/3
(D) 65/81
(E) 71/81
The sharpshooter hits the target once in 3 shots. Hence, the probability of hitting the target is 1/3. The
probability of not hitting the target is 1 – 1/3 = 2/3.
Now, (the probability of not hitting the target even once in 4 shots) + (the probability of hitting at least once
in 4 shots) equals 1, because these are the only possible cases.
Hence, the probability of hitting the target at least once in 4 shots is
1 – (the probability of not hitting even once in 4 shots)
The probability of not hitting in the 4 chances is
2/3.2/3.2/3.2/3=16/81 Now, 1 – 16/81 = 65/81. The answer is (D).
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