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A man can hit a target once in 4 shots

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A man can hit a target once in 4 shots [#permalink] New post 07 Nov 2010, 23:57
00:00
A
B
C
D
E

Difficulty:

  5% (low)

Question Stats:

44% (01:01) correct 56% (00:03) wrong based on 9 sessions
A man can hit a target once in 4 shots. If he fires 4 shots in succession, what is the probability that he will hit his target?

A) 1
B) 1/256.
C) 81/256
D) 175/256
E) 144/256

Official answer is D. My question is why cannot it be A?
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Re: Hit a target - Probability. [#permalink] New post 08 Nov 2010, 00:07
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soundofmusic wrote:
I have a question,

A man can hit a target once in 4 shots. If he fires 4 shots in succession, what is the probability that he will hit his target?

A) 1.
B) 1/256.
C) 81/256.
D) 175/256
E) 144/256

Official answer is D.My question is why cannot it be A?


The wording of the question is not the one you'll see on GMAT but anyway: "A man can hit a target once in 4 shots" means that the probability of hitting the target is 1/4, thus the probability of missing the target is 1-1/4=3/4.

The probability that he will hit a target while shooting 4 times is basically the probability that he will hit at least once, so 1-(the probability that he will miss all 4 times): P=1-(\frac{3}{4})^4=\frac{175}{256}.

Answer: D.
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Re: Hit a target - Probability. [#permalink] New post 08 Nov 2010, 00:19
Thanx Bunuel....i guess the wording just got me confused.
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Probability question [#permalink] New post 08 Jan 2012, 02:21
Here's the question

A man can hit a target once in 4 shots. If he fires 4 shots in succession, what is the probability that he will hit his target?

A. 1

B. 255/256

C. 175/256

D. 1/4

E. 1/2

I do know how to get it using the 1-(No hits). However, I can't for the life of me get it by adding probabilities together.

This is what I am doing

P(One hit) = (1/4)*(3/4)^3 = 27/256
P(Two hits)=(1/4)^2 * (3/4)^2= 9/256
P(Three hits)=(1/4)^3 * (3/4)= 3/256
P(Four hits) = (1/4)^4 = 1/256

Adding them all together would yield 40/256 which is wrong.

I am missing something.

Thanks in advance.

Source is 4gmat.com
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Re: Probability question [#permalink] New post 08 Jan 2012, 13:43
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Try a bernoulli trial here.
P(k)=nCk*p^k*(1-p)^(n-k) where n is the number of shots k is the number of hits p is probability of a hit

set k=0 and you get

P(k)=4C0*(3/4)^4=3^4/4^4=81/256

1-P(k)=(256-81)/256=175/256

Hope this helped

Last edited by joeshmo on 08 Jan 2012, 14:18, edited 1 time in total.
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Re: Probability question [#permalink] New post 08 Jan 2012, 14:04
joeshmo wrote:
Try a bernoulli trial here.
P(k)=nCk*p^n*(1-p)^(n-k) where n is the number of shots k is the number of hits p is probability of a hit

set k=0 and you get

P(k)=4C0*(3/4)^4=3^4/4^4=81/256

1-P(k)=(256-81)/256=175/256

Hope this helped


Thanks Joe.

This does lead to the right answer, but I am still missing how to add them all up. It's the classical, albeit slower, way of going about these problem.s
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Re: Probability question [#permalink] New post 08 Jan 2012, 14:18
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You are missing the nCk multiplier portion so for 1 hit it would be 4C1 for 2 hits it would be 4C2 for 3 hits it would be 4C3

So 1 hit

P(1 hit)=4C3*(1/4)*(3/4)^3
P(2 hit)=4C2*(1/4)^2*(3/4)^2
P(3 hit)=4C1*(1/4)^3*(3/4)^1
P(4 hit)=4C0*(1/4)^4

P(1)=108/256
P(2)=54/256
P(3)=12/256
P(4)=1/256
P(1)+P(2)+P(3)+P(4)=175/256
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Re: Probability question [#permalink] New post 08 Jan 2012, 19:35
Probability = 1st hit + 2nd hit + 3rd hit + 4th hit
1st = (1/4)(3/4)^3 X 4C3
2nd = (1/4)^2(3/4)^2 x 4C2
3rd = (1/4)^3(3/4)^1 x 4C1
4th = (1/4)^4

Total = 175/256

Total =
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Re: Probability question [#permalink] New post 09 Jan 2012, 04:43
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Lstadt wrote:
Here's the question

A man can hit a target once in 4 shots. If he fires 4 shots in succession, what is the probability that he will hit his target?

A. 1

B. 255/256

C. 175/256

D. 1/4

E. 1/2

I do know how to get it using the 1-(No hits). However, I can't for the life of me get it by adding probabilities together.

This is what I am doing

P(One hit) = (1/4)*(3/4)^3 = 27/256
P(Two hits)=(1/4)^2 * (3/4)^2= 9/256
P(Three hits)=(1/4)^3 * (3/4)= 3/256
P(Four hits) = (1/4)^4 = 1/256

Adding them all together would yield 40/256 which is wrong.

I am missing something.

Thanks in advance.

Source is 4gmat.com


There are various ways of getting the answer. Here are a few of them.

Method 1: (Simplest) To hit the target, you need to hit it at least once in the 4 shots. You would not have hit the target, if you do not hit it in any of the 4 shots.
Probability of not hitting the target in any of the 4 shots = (3/4)*(3/4)*(3/4)*(3/4) = 81/256
Probability of hitting the target at least once = 1 - 81/256 = 175/256

Method 2: You hit the target if you hit it at least once. Say, if you hit the target on the first shot, you are done no matter what you do on the rest of the 3 shots. If you do not hit the target on the first shot, but hit it on the second shot, again, you are done after that... etc
Probability of hitting the target = (1/4) + (3/4)*(1/4) + (3/4)*(3/4)*(1/4) + (3/4)*(3/4)*(3/4)*(1/4) = (1*64 + 3*16 + 9*4 + 27)/256 = 175/256

Method 3: Say H = Hit and M = Miss
Probability of hitting the target once = (1/4)*(3/4)*(3/4)*(3/4) * 4!/3! = 27*4/256
Why do we multiply by 4!/3!? because 1 Hit and 3 Misses can be arranged in 4 ways: HMMM or MHMM or MMHM or MMMH. You arrange 4 things out of which 3 are identical in 4 ways.

Probability of hitting the target twice = (1/4)*(1/4)*(3/4)*(3/4) * 4!/2!*2! = 54/256
Why do we multiply by 4!/2!*2!? Same logic as above. You arrange 4 things out of which 2 pairs are identical in 4!/2!*2! ways.

Probability of hitting the target thrice = (1/4)*(1/4)*(1/4)*(3/4) * 4!/3! = 12/256

Probability of hitting the target four times = (1/4)*(1/4)*(1/4)*(1/4) = 1/256

Add them all up. You will get 175/256
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Re: Probability question [#permalink] New post 09 Jan 2012, 10:00
Thank you All especially Karishma.
Re: Probability question   [#permalink] 09 Jan 2012, 10:00
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