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Three snipers shoot a certain target. Their probabilities of hitting

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Bunuel
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Three snipers shoot a certain target. Their probabilities of hitting [#permalink] New post   09 Jan 2021, 02:16
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Three snipers shoot a certain target. Their probabilities of hitting the target are 0.9, 0.7, and 0.5 respectively. The snipers make one salvo. What is the probability that exactly one sniper missed?

A. 0.015
B. 0.185
C. 0.485
D. 0.515
E. 0.985

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The snipers shoot a certain target. Their probabilities of hitting the target are 0.9, 0.7, and 0.5 respectively. The snipers make one salvo. What is the probability:

1. that exactly one sniper missed?
2. that exactly one sniper hit?
3. that exactly two snipers missed?
4. that exactly two snipers hit?
ShowHide Answer
Official Answer
C
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TarunKumar1234
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Re: Three snipers shoot a certain target. Their probabilities of hitting [#permalink] New post   09 Jan 2021, 03:01
Three snipers shoot a certain target. Their probabilities of hitting the target are 0.9, 0.7, and 0.5 respectively. The snipers make one salvo. What is the probability that exactly one sniper missed?

A= 0.9, B= 0.7 and C= 0.5

Total Probability= Only A miss + Only B miss+ Only C miss = (1-0.9)*0.7*0.5 + 0.9*(1-0.7)*0.5 + 0.9*0.7*(1-0.5)
= 0.035+0.135+0.315= 0.485

So, I think C. :)
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Re: Three snipers shoot a certain target. Their probabilities of hitting [#permalink] New post   09 Jan 2021, 04:20
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TarunKumar1234 wrote:
Three snipers shoot a certain target. Their probabilities of hitting the target are 0.9, 0.7, and 0.5 respectively. The snipers make one salvo. What is the probability that exactly one sniper missed?

A= 0.9, B= 0.7 and C= 0.5

Total Probability= Only A miss + Only B miss+ Only C miss = (1-0.9)*0.7*0.5 + 0.9*(1-0.7)*0.5 + 0.9*0.7*(1-0.5)
= 0.035+0.135+0.315= 0.485

So, I think C. :)
Bunuel wrote:
Three snipers shoot a certain target. Their probabilities of hitting the target are 0.9, 0.7, and 0.5 respectively. The snipers make one salvo. What is the probability that exactly one sniper missed?

A. 0.015
B. 0.185
C. 0.485
D. 0.515
E. 0.985

Show Spoiler
The snipers shoot a certain target. Their probabilities of hitting the target are 0.9, 0.7, and 0.5 respectively. The snipers make one salvo. What is the probability:

1. that exactly one sniper missed?
2. that exactly one sniper hit?
3. that exactly two snipers missed?
4. that exactly two snipers hit?


It is 'C' indeed.
Probability
Soldier 1 - hit-0.9 / miss-0.1
Soldier 2 - hit-0.7 / miss-0.3
Soldier 3 - hit-0.5 / miss-0.5

Probability that exactly one sniper missed -
(1) miss and (2)(3) hit = (0.1)(0.7)(0.5)
OR (+)
(2) miss and (1)(3) hit = (0.9)(0.3)(0.5)
OR (+)
(3) miss and (1)(2) hit = (0.9)(0.7)(0.5)
Answer = option C. 0.485
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Re: Three snipers shoot a certain target. Their probabilities of hitting [#permalink] New post   09 Jan 2021, 08:08
I think the answer is C. Here is my two cents

Let 3 shoots A, B and C.

P(A)=0.9,P(A′)=1−0.9=0.1
P(B)=0.7,P(B′)=1−0.7=0.3
P(C)=0.5,P(C′)=1−0.5=0.5
Probability of exactly one sniper missed = P(A)×P(B)×P(C′)+P(B)×P(C)×P(A′)+P(C)×P(A)×P(B′)
=0.9×0.7×0.5+0.7×0.5×0.1+0.5×0.9×0.3
=0.315+0.035+0.135=0.485
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Re: Three snipers shoot a certain target. Their probabilities of hitting [#permalink] New post   17 Jun 2022, 04:41
Let the snipers with hitting probabilities of 0.9, 0.7, and 0.5 be A, B, and C, respectively.
Thus, the probability of missing will be 0.1, 0.3, and 0.5, respectively.

There are 3 cases:
(i) A missed: Probability = 0.1*0.7*0.5 = 0.035
(ii)B missed: Probability = 0.9*0.3*0.5 = 0.135
(iii)C missed: Probability = 0.9*0.7*0.5 = 0.315

Thus, probability of any one missing will be 0.035 + 0.135 + 0.315 = 0.485

Thus, the correct option is C.
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Re: Three snipers shoot a certain target. Their probabilities of hitting [#permalink] New post   17 Jun 2022, 05:10
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This question can be answered by some ball-parking.

Notice that the average hit rate is .7, so the average miss rate is 0.3.

It's not legitimate to then multiply these in hopes of arriving at the precise answer but will be a reasonable approximation.

The probability of two hits and one miss is .7^2 * .3 multiplied by 3, since any one of the three could miss:

3*.3=.9
.7^2 = .49

.9*.49 is certainly less than D and much greater than B, so C must be the answer

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Re: Three snipers shoot a certain target. Their probabilities of hitting [#permalink]
17 Jun 2022, 05:10
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