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# 3 archers

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Director
Joined: 07 Jun 2004
Posts: 613
Location: PA
Followers: 3

Kudos [?]: 522 [0], given: 22

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08 Nov 2010, 03:51
00:00

Difficulty:

35% (medium)

Question Stats:

63% (01:41) correct 38% (01:15) wrong based on 16 sessions

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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: 33090
Followers: 5778

Kudos [?]: 70923 [1] , given: 9857

Re: 3 archers [#permalink]

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08 Nov 2010, 04:14
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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, 04:14
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# 3 archers

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