Probability Of Shots

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manish.jaipatna: Junior | Next Rank: 30 Posts; Posts: 21; Joined: Tue Jul 20, 2010 11:54 pmFollowed by: 1 member

by manish.jaipatna » Wed Mar 23, 2011 5:38 pm

Hi All,

Can you please help me in solving below probability question.

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?

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

Thanks
Manish

Thanks
Manish
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Source: Beat The GMAT — Problem Solving | Wed Mar 23, 2011 5:38 pm

ajmoney09: Master | Next Rank: 500 Posts; Posts: 110; Joined: Sun May 25, 2008 3:45 pm

by ajmoney09 » Wed Mar 23, 2011 5:58 pm

I said A.

1 out of ever 4...shoots 4 he is going to hit it once.

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manish.jaipatna: Junior | Next Rank: 30 Posts; Posts: 21; Joined: Tue Jul 20, 2010 11:54 pmFollowed by: 1 member

by manish.jaipatna » Wed Mar 23, 2011 6:03 pm

ajmoney09,

Can you explain me in more detail as i am very weak in probability questions. Also if you have any site to refer for probability concepts then that would be great.

Thanks
Manish

Thanks
Manish
"The Purpose Of Life Is A Life Of Purpose"

"Ruk Jaana Nahi Kanhi Tu Haarke"

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coolly01: Junior | Next Rank: 30 Posts; Posts: 29; Joined: Sun Feb 20, 2011 11:03 pm

by coolly01 » Wed Mar 23, 2011 6:04 pm

I pick B
1 probability in 4 shot (average )

=> he has 1/4 probability to hit the target in one shot
Then in 4 successive shots he has : 1/4 x 1/4 x 1/4 x 1/4 probability to hit the target => 1/256
What is the OA?

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srcc25anu: Master | Next Rank: 500 Posts; Posts: 423; Joined: Fri Jun 11, 2010 7:59 amFollowed by: 2 members; Location: Seattle, WA; Thanked: 86 times

by srcc25anu » Wed Mar 23, 2011 6:40 pm

P(hit) = 1/4
P(no hit) = 3/4
find the prob that he doesnt hit the target in 4 shots thats equal to (3/4)^4
subtract this from 1 to get the desired probability (that he will hit the target)
therefore 1-81/256 = 175/256
IMO D

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manish.jaipatna: Junior | Next Rank: 30 Posts; Posts: 21; Joined: Tue Jul 20, 2010 11:54 pmFollowed by: 1 member

by manish.jaipatna » Wed Mar 23, 2011 6:52 pm

srcc25anu,

You are correct. Thanks a lot. But i am not always getting an approach to answer such question

. Another such question is

We have to choose four numbers between 1 to 10. What is the probability of getting 2 & 3 from the chosen four numbers?

Thanks
Manish
"The Purpose Of Life Is A Life Of Purpose"

"Ruk Jaana Nahi Kanhi Tu Haarke"

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srcc25anu: Master | Next Rank: 500 Posts; Posts: 423; Joined: Fri Jun 11, 2010 7:59 amFollowed by: 2 members; Location: Seattle, WA; Thanked: 86 times

by srcc25anu » Wed Mar 23, 2011 7:14 pm

we find the prob of not selecting a 2&3 in 8c4 / 10c4 = 1/3
subtract this from 1 to get the desired probability. that should be 1-1/3 = 2/3

manish.jaipatna wrote:srcc25anu,

We have to choose four numbers between 1 to 10. What is the probability of getting 2 & 3 from the chosen four numbers?

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manpsingh87: Master | Next Rank: 500 Posts; Posts: 436; Joined: Tue Feb 08, 2011 3:07 amFollowed by: 6 members; Thanked: 72 times

by manpsingh87 » Wed Mar 23, 2011 7:37 pm

manish.jaipatna wrote:Hi All,

Can you please help me in solving below probability question.

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?

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

Thanks
Manish

probability of hitting the target=1/4;
probability of not hitting the target=3/4;

H- Hit the target
M- Miss the target

Required probability = H+MH+MMH+MMMH
=1/4+3/4*(1/4)+(3/4)*(3/4)*(1/4)+(3/4)*(3/4)*(3/4)*(1/4);
=64+48+36+27/256
=175/276

hence D

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