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Shooting Competition

by harsh.champ » Wed Feb 17, 2010 6:02 am
In a shooting competition, the probability of hitting the target by A is 2/5, by B is 2/3 and by C is 3/5. If all of them fire independently at the same target, calculate the probability that only one of them will hit the target.

A.1/2
B.1/3
C.1/4
D.1/5
E.1/6
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by Osirus@VeritasPrep » Wed Feb 17, 2010 7:02 am
Is this problem missing something? Who is the "they"? go back to complore and recopy this. Thanks in advance.
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by money9111 » Wed Feb 17, 2010 7:55 am
are the variables A, B, and C distances? you said probabilit yof hitting a target BY..
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by ajith » Wed Feb 17, 2010 9:08 am
harsh.champ wrote:In a shooting competition, the probability of hitting the target by A is 2/5, by B is 2/3 and by C is 3/5. If all of them fire independently at the same target, calculate the probability that only one of them will hit the target.

A.1/2
B.1/3
C.1/4
D.1/5
E.1/6
3 cases A hits B C misses, prob = 2/5*1/3*2/5 = 4/75
B hits A C misses, prob = 3/5*2/3*2/5 = 12/75
C hits A B misses = 3/5*1/3*3/5 = 9/75

total = 25/75 = 1/3
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by shashank.ism » Wed Feb 17, 2010 9:22 am
money9111 wrote:are the variables A, B, and C distances? you said probabilit yof hitting a target BY..
Its clearly written probability of hitting the target by A is 2/5, by B is 2/3 and by C is 3/5.. how can "by" refer a distance. A, B and C are not variables but they are referring to a person(either male or female or ...)
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by shashank.ism » Wed Feb 17, 2010 9:27 am
harsh.champ wrote:In a shooting competition, the probability of hitting the target by A is 2/5, by B is 2/3 and by C is 3/5. If all of them fire independently at the same target, calculate the probability that only one of them will hit the target.

A.1/2
B.1/3
C.1/4
D.1/5
E.1/6
P(A) = 2/5, P(B) =2/3, P(C)= 3/5 -----> P(A') = 3/5, P(B') =1/3, P(C')= 2/5
P(only one hit target) = P(A')P(B')P(C)+P(A')P(B)P(C')+P(A)P(B')P(C') = 3/5 x 1/3 x 3/5 + 3/5 x 2/3 x 2/5 + 2/5 x 1/3 x 2/5 = 25/75 =1[spoiler]/3 Ans B[/spoiler]
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by money9111 » Wed Feb 17, 2010 9:53 am
@ shashank.ism - i'm going to disagree with you here... by saying that the probability of hitting the target BY A is 2/5... could and in my opinion does in fact mean a distance.

i think you... well Harsh.Champ meant to type "...hitting the Target A is 2/5" and not "...hitting the target by A is 2/5" do you see the difference?
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by BSDesis » Wed Feb 17, 2010 2:45 pm
money9111 wrote:@ shashank.ism - i'm going to disagree with you here... by saying that the probability of hitting the target BY A is 2/5... could and in my opinion does in fact mean a distance.

i think you... well Harsh.Champ meant to type "...hitting the Target A is 2/5" and not "...hitting the target by A is 2/5" do you see the difference?
Money, substitute A for a person's name and I think you'll get where shashank is coming from.

If A=Money9111, the original stem reads,
"In a shooting competition, the probability of hitting the target by Money9111 is 2/5..."

It could be rewritten so that "In a shooting competition, the probability of A (Money9111) hitting the target is 2/5" is less ambiguous, but A musn't be a distance I do not think.
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by harsh.champ » Wed Feb 17, 2010 10:00 pm
osirus0830 wrote:Is this problem missing something? Who is the "they"? go back to complore and recopy this. Thanks in advance.
Hey osirus,
I couldn't get ur doubt-"They".Are you finding any problem with the wording of the problem?
Anyways this problem is not from complore. It is a very common type of probability problem,I guess and can be found in numerous resources.I don't know if complore is one of them.Anyways,do let me know your doubt about the question.
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by harsh.champ » Wed Feb 17, 2010 10:02 pm
money9111 wrote:are the variables A, B, and C distances? you said probabilit yof hitting a target BY..
No,A,B,C are not distances.They are 3 different persons.[You can change A,B,C to Tom,Dick and Harry to get a better clarity of the question]Probability of hitting the target by A means "if the 1st person hits the target" not "the distance y which he/she hits it".
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by harsh.champ » Wed Feb 17, 2010 10:13 pm
harsh.champ wrote:In a shooting competition, the probability of hitting the target by A is 2/5, by B is 2/3 and by C is 3/5. If all of them fire independently at the same target, calculate the probability that only one of them will hit the target.

A.1/2
B.1/3
C.1/4
D.1/5
E.1/6
Well guys ,I would just post the solution now so that nobody has any other doubts.
I guess if not specified otherwise,we take the events to be mutually exclusive.
Soln:- P(only one of them hit the target ) = P(A hits,B misses,C misses) + P(B hits,A misses,C misses) +
P(C hits,B misses,A misses)
=(2/5 x 1/3 x 2/5) + (2/3 x 3/5 x 2/5) + (3/5 x 3/5 x 1/3)
=[spoiler]25/75 = 1/3 B[/spoiler]

I guess the soln. is clear now.
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by shashank.ism » Thu Feb 18, 2010 6:28 am
BSDesis wrote:
money9111 wrote:@ shashank.ism - i'm going to disagree with you here... by saying that the probability of hitting the target BY A is 2/5... could and in my opinion does in fact mean a distance.

i think you... well Harsh.Champ meant to type "...hitting the Target A is 2/5" and not "...hitting the target by A is 2/5" do you see the difference?
Money, substitute A for a person's name and I think you'll get where shashank is coming from.

If A=Money9111, the original stem reads,
"In a shooting competition, the probability of hitting the target by Money9111 is 2/5..."

It could be rewritten so that "In a shooting competition, the probability of A (Money9111) hitting the target is 2/5" is less ambiguous, but A musn't be a distance I do not think.
Thanks BSDesis to clear the concept by giving some real life examples..
I meant the same that A,B, C are the persons Just replace A,B,C by any persons name and you will feel that it is actually representing a person..
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