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Source: — Problem Solving |
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by cans » Sun Jun 05, 2011 4:12 pm
prob that a man can hit = 1/4 (once in 4 shots)
and thus prob that he misses = 1 - 1/4 = 3/4
4 shots fired. let H represents hit and M represents miss
this p(Hit the target) = P(H) + P(MH) + P(MMH) + P(MMMH)
= 1/4 + (3/4)*(1/4) + (3/4)(3/4)(1/4) + (3/4)(3/4)(3/4)(1/4)
1/4 + 3/16 + 9/64 + 27/256 = 175/256

IMO D
P(H) + P(MH) + P(MMH) + P(MMMH) (we get this as if he hits first, he won't need to fire any other shot. MH - misses first and hits in the 2nd. MMH - misses first 2 shots and hits in 3rd. MMMH - misses first 3 shots and hits in the 4th.)
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by Brent@GMATPrepNow » Sun Jun 05, 2011 5:31 pm
vinitrathi1 wrote:Q.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?

1. 1
2. 1/256
3. 81/256
4. 175/256
5. 144/256
Another way to solve this is to say that P(hit target at least once in 4 tries) = 1 - P(hit target zero times in 4 tries)

P(hit target zero times in 4 tries)=(3/4)(3/4)(3/4)(3/4)=81/256

So, P(hit target at least once in 4 tries) = 1 - 81/256
=175/256

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by pemdas » Sun Jun 05, 2011 5:54 pm
Hi Brent, don't we violate independence of events-Hit/Miss here?
P(hit target zero times in 4 tries)=(3/4)(3/4)(3/4)(3/4)=81/256
with each successive try the probability must change, BUT you keep it absolute 3/4 (Miss) at every successive try

i ask to reveal the source
i'm not sure though - OA isn't in line with what i've obtained (in spoiler) it's close to ans. 4 but deviation of 2% (175/256=68%, my calc is 70%)
[spoiler]
i would seek 3/4*2/3*1/2*1 first try
next tries 1/4*2/3*1/2*1, 3/4*1/3*1/2*1 and 3/4*2/3*1/2*1

1/4+1/12+1/8+1/4=(6+2+3+6)/24=17/24[/spoiler]
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by lunarpower » Wed Jun 08, 2011 2:15 am
this is another problem on which the issue is bad wording -- "a man can hit a target once in 4 shots" is ambiguous.
pemdas is interpreting this statement as "out of exactly 4 shots, exactly 1 will hit the target". brent is reading it as "he has a 1/4 chance of hitting the target, regardless of what has already happened."

brent's interpretation is more in line with my own intuition about what the writers *meant* to say, but the question is genuinely ambiguous.

in any case, though --
i can see that some posters (pemdas in particular) don't know the source of the problem.

i'll ask the same question i've asked about a thousand times:
IF YOU DON'T KNOW THE SOURCE, WHY ARE YOU USING THE PROBLEM?

do not use problems from unknown third-party sources -- most of them are TERRIBLE, and many (if not most) of them will basically cause you to un-learn, or get confused about, the things that you're supposed to learn.
Ron has been teaching various standardized tests for 20 years.

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by pemdas » Wed Jun 08, 2011 10:10 am
Ron, thanks for resolving my doubt.
>> why with so many educated gmatters, we are still having an unknown source issue?
i doubt any one with much less education/schooling than us at BTG, ever buys beans at the groceries store without the origin, expiry codes
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by lunarpower » Thu Jun 09, 2011 1:09 am
pemdas wrote: i doubt any one with much less education/schooling than us at BTG, ever buys beans at the groceries store without the origin, expiry codes
well done.
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