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# In a shooting competition, probability of A hitting target

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Director
Joined: 24 Oct 2005
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In a shooting competition, probability of A hitting target [#permalink]

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28 Oct 2005, 04:12
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100% (02:53) correct 0% (00:00) wrong based on 4 sessions

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In a shooting competition, probability of A hitting target is 2/5, by B is 2/3 and C is 3/5. If all of them fire independently, what is the probability that only one will hit target?
A) 4/25
B) 1/3
C) 3/25
4) 2/3
Director
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28 Oct 2005, 04:51
probability of A hitting target is 2/5, by B is 2/3 and C is 3/5

a: 2/5 * 1/3 * 2/5 = 4/75
b: 3/5 * 2/3 * 2/5 = 12/75
c: 3/5 * 1/3 * 3/5 = 9/75
-------------------------------
a+b+c=25/75=1/3
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Director
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28 Oct 2005, 04:52
(2/5x1/3x2/5)+(2/3x3/5x2/5)+(3/5x1/3x2/5)=22/75
My answer doesn,t seem to mach the answer choices.Guess it is B)
Director
Joined: 14 Sep 2005
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Location: South Korea
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28 Oct 2005, 04:58
BG wrote:
(2/5x1/3x2/5)+(2/3x3/5x2/5)+(3/5x1/3x2/5)=22/75
My answer doesn,t seem to mach the answer choices.Guess it is B)

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Auge um Auge, Zahn um Zahn !

Director
Joined: 24 Oct 2005
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Location: London
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28 Oct 2005, 05:02

P(A) hitting target = 2/5. Therefore P(A) not hitting = 3/5
P(B) = 2/3 . Therefore P(not B) = 1/3
P(C) = 3/5 . Therefore P(not C) = 2/3

P (only A) means A hits and B , C dont .
ie. 2/5 * 1/3 * 2/3
P (only B) means B hits and A , C dont .
ie. 2/3 * 3/5 * 2/3
P (only C) means C hits and A, B dont .
ie. 3/5 * 3/5 * 1/3

Probability = P(only A) or p(only B) or p(only C)
ie. (2/5 * 1/3 * 2/3) + (2/3 * 3/5 * 2/3) + (3/5 * 3/5 * 1/3) = 1/3

Hence B
28 Oct 2005, 05:02
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# In a shooting competition, probability of A hitting target

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