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3 archers

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3 archers [#permalink] New post 08 Nov 2010, 02:51
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Difficulty:

  35% (medium)

Question Stats:

71% (01:41) correct 29% (01:21) wrong based on 14 sessions
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
[Reveal] Spoiler: OA

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Re: 3 archers [#permalink] New post 08 Nov 2010, 03:14
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rxs0005 wrote:
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


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.
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Re: 3 archers   [#permalink] 08 Nov 2010, 03:14
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