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3 archers

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Director
Joined: 07 Jun 2004
Posts: 614
Location: PA
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Kudos [?]: 290 [0], given: 22

3 archers [#permalink]  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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Math Expert
Joined: 02 Sep 2009
Posts: 27471
Followers: 4312

Kudos [?]: 42229 [1] , given: 5968

Re: 3 archers [#permalink]  08 Nov 2010, 03:14
1
KUDOS
Expert's post
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 $$P=(\frac{1}{3})^3=\frac{1}{27}$$.

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