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In a shooting competition, probability of A hitting target is 2/5, by

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Math Expert
Joined: 02 Sep 2009
Posts: 58320
In a shooting competition, probability of A hitting target is 2/5, by  [#permalink]

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13 Nov 2018, 10:48
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Difficulty:

25% (medium)

Question Stats:

75% (01:59) correct 25% (01:53) wrong based on 82 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) 3/25
(B) 4/25
(C) 1/3
(D) 2/3
(E) 3/4

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Re: In a shooting competition, probability of A hitting target is 2/5, by  [#permalink]

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13 Nov 2018, 11:07
1
2
P(x) = probability of hitting target, P(x") = probability of not hitting target = 1 - P(x), using this:

P(A) = 2/5, so, P(A") = 3/5
P(B) = 2/3, so, P(B") = 1/3
P(C) = 3/5, so, P(C") = 2/5

OR = plus, and = *

We need: P(A hits & B doesnt & C doesnt) OR P(A doesnt & B hits & C doesnt) OR P(A doesnt & B doesnt & C hits)
which is: P(A)(B")(C") + P(A")P(B)P(C") + P(A")P(B")P(C)

= $$(2/5)*(1/3)*(2/5) + (3/5)*(2/3)*(2/5) + (3/5)*(1/3)*(3/5)$$

= $$\frac{(4+12+9)}{75}$$

= 25/75 = 1/3

Option C
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Re: In a shooting competition, probability of A hitting target is 2/5, by  [#permalink]

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14 Nov 2018, 02:10
Bunuel wrote:
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) 3/25
(B) 4/25
(C) 1/3
(D) 2/3
(E) 3/4

P (A)= 2/5 , Pn(A)= 3/5
P(B)= 2/3, Pn(B)=1/3
P(C)= 3/5, Pn(C)=2/5

only one will hit Probability=
{2/5*1/3*2/5}+{3/5*2/3*2/5}+{3/5*1/3*3/5}

= 25/75
=1/3 option C
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Re: In a shooting competition, probability of A hitting target is 2/5, by  [#permalink]

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27 Jan 2019, 21:04
Bunuel wrote:
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) 3/25
(B) 4/25
(C) 1/3
(D) 2/3
(E) 3/4

P(A hits and B,C miss) = 2/5 x 1/3 x 2/5 = 4/75

P(B hits and A, C miss) = 3/5 x 2/3 x 2/5 = 12/75

P(C hits and A, B miss) = 3/5 x 1/3 x 3/5 = 9/75

So the total probability is 4/75 + 12/75 + 9/75 = 25/75 = 1/3.

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Re: In a shooting competition, probability of A hitting target is 2/5, by   [#permalink] 27 Jan 2019, 21:04
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