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08 Nov 2010, 03:51
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A
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
45%
(medium)
Question Stats:
66% (01:28) correct
34%
(02:04)
wrong
based on 231
sessions
History
Date
Time
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Not Attempted Yet
Three archers each have an equal chance of hitting a target, and if only two of the three shoot the likelihood of the two hitting the target is 4/9 . What is the likelihood of all three men missing the target?
(A) 1/27
(B) 16/81
(C) 8/27
(D) 19/27
(E) 26/27
(A) 1/27
(B) 16/81
(C) 8/27
(D) 19/27
(E) 26/27
08 Nov 2010, 04:14
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rxs0005
Let the probability of an archer hitting a target be \(x\). Then the probability of two hitting the target will be \(P=x*x=\frac{4}{9}\) --> \(x=\frac{2}{3}\), so the probability of an archer missing the target will be \(P=1-\frac{2}{3}=\frac{1}{3}\).
The probability of all three men missing the target will be \(P=(\frac{1}{3})^3=\frac{1}{27}\).
Answer: A.
General Discussion
21 May 2019, 19:26
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rxs0005
Let p = the likelihood (i.e., probability) that one hits the target. From the information provided, we we have:
p^2 = 4/9
p = 2/3
Therefore, the probability that one archer doesn’t hit the target is ⅓, and hence the probability that all three don’t hit the target is 1/3 x 1/3 x 1/3 = 1/27.
Answer: A
