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08 Nov 2010, 00:57
Kudos
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D
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
65%
(hard)
Question Stats:
56% (01:34) correct
44%
(01:14)
wrong
based on 471
sessions
History
Date
Time
Result
Not Attempted Yet
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
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?
09 Jan 2012, 05:43
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Bookmarks
Lstadt
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
08 Nov 2010, 01:07
Kudos
Bookmarks
soundofmusic
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.
