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4 shots in succession

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4 shots in succession

by sanju09 » Tue Apr 05, 2011 4:32 am
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) 1/256
(B) 81/256
(C) 144/256
(D) 175/256
(E) 1
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by vineeshp » Tue Apr 05, 2011 4:36 am
E?
Since he hits once in 4 and 4 are fired, probability is 1.

If this is wrong, I am going back to my basics :D
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by sanju09 » Tue Apr 05, 2011 4:37 am
vineeshp wrote:E?
Since he hits once in 4 and 4 are fired, probability is 1.

If this is wrong, I am going back to my basics :D
happy journey to your basics
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by vineeshp » Tue Apr 05, 2011 6:26 am
Can you please post the OA and solution? If you don wanna post here, it would be great if you can message it to me.
Thanks.
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by rohu27 » Tue Apr 05, 2011 7:07 am
let me try
the probability tht the man hits the target is 1/4
now he is shooting 4 times in succession at the target so if he hits target for 1st time he is succesful,if he hits both 1s and 2nd time he his succesful, 1st 2nd and 3rd also succesful, all 4 also he is successful.
so we can find the probabilty whn he doesnt hit the target and subtact it frm 1 (a la atleast question)
probailty tht he doesnt hit the target for all the 4 shots - 3/4*3/4*3/4*3/4=81/256
1-81/256=175/256
option D
not sure if its right though, sanju can confirm.
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by manpsingh87 » Tue Apr 05, 2011 7:07 am
sanju09 wrote: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) 1/256
(B) 81/256
(C) 144/256
(D) 175/256
(E) 1
as he hit once in 4 shots therefore probability of hitting the target is 1/4.
and probability of not hitting the target = 1-1/4=3/4;
probability of hitting the target in 4 successive shorts= 1- (probability of not hitting the target in four shots)
1-(3/4)*(3/4)*(3/4)*(3/4)= 175/256...!!!! therefore D
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by vineeshp » Tue Apr 05, 2011 7:25 am
Thanks guys. Your solution is correct I guess.
I was trying to solve problems at a fast rate in office without my colleagues seeing. Just blabbered something.
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by tpr-becky » Tue Apr 05, 2011 8:17 am
This concept of this problem is probabilty:

Probability is either based on "and"; "or" or "at least" on the GMAT - since we want to know the probability that he hits is target in four shots he could hit his target 1,2,3, or 4 times - therefore what we really want is the probabilty he hits his target AT LEAST once.

At least probability is found by the subtracting the probability that he does not hit his target from 1.

The probablity he does not hit is target is 3/4 each time - thus it is 3^4/4^4 = 81/256

1-81/256 = 175/256 so that is the answer.
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by force5 » Wed Apr 06, 2011 3:25 am
Hi sanju09... can i ask you the source of this problem?? Didn't like the question....
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by Obsessed » Wed Apr 06, 2011 10:13 am
Thanks Becky for giving an explaination to the answer.
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by manpsingh87 » Wed Apr 06, 2011 7:30 pm
hi everyone, as we've already solved the question, let's see an alternate way of solving it,i.e. how can we reach at 175 ways of hitting the target out of possible 256.

here four different cases are possible:
1) 1 shot hit the target.
2) 2 shots hit the target.
3) 3 shots hit the target.
4) 4 shots hit the target.

1) x x x x out of the 4 shots one shot can be selected in 4C1 ways, therefore required probability is 4C1*(1/4)*(3/4)*(3/4)*(3/4)=108/256;
2) x x x x out of the 4 shots any two shots that will hit the target can be selected in 4C2 ways, therefore required probability is 4C2*(1/4)*(1/4)*(3/4)*(3/4)=54/256;
3) x x x x out of 4 shots 3 shots can be selected in 4C3 ways, therefore required probability is
4C3*(1/4)*(1/4)*(1/4)*(3/4)=12/256;

4) probability when all the shots hit the target =(1/4)*(1/4)*(1/4)*(1/4)=1/256;

hence required probability is 1)+2)+3)+4)=(108+54+12+1)/256;=175/256
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