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probability:x,y,z

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Director
Joined: 25 Oct 2008
Posts: 610
Location: Kolkata,India
Followers: 9

Kudos [?]: 288 [0], given: 100

probability:x,y,z [#permalink]  17 Aug 2009, 19:18
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Question Stats:

100% (01:46) correct 0% (00:00) wrong based on 1 sessions
x,y and z each try independantly to solv a problem.If their individual probabilities for success are 1/4,1/2 and 5/8 resp.,what is the probability that x and y but not z will solve the problem?

Guys this is my approach where am I going wrong?
Z does not solve the problem=only X solves it+ only Y solves it + both x and Y solve it.

=1/4+1/2+1/4*1/2
=1/4+1/2+1/8
=7/8...which is the wrong ans...
_________________
Senior Manager
Joined: 17 Mar 2009
Posts: 310
Followers: 6

Kudos [?]: 269 [0], given: 22

Re: probability:x,y,z [#permalink]  17 Aug 2009, 19:38
tejal777 wrote:
x,y and z each try independantly to solv a problem.If their individual probabilities for success are 1/4,1/2 and 5/8 resp.,what is the probability that x and y but not z will solve the problem?

Guys this is my approach where am I going wrong?
Z does not solve the problem=only X solves it+ only Y solves it + both x and Y solve it.

=1/4+1/2+1/4*1/2
=1/4+1/2+1/8
=7/8...which is the wrong ans...

you have answered to the question . "What is the probability of x OR y solving the problem?"

you must include z's probability of not solving that is (1-5/8) = 3/8
1/4 * 1/2 * 3/8 = 3/64, is this is the answer?
Director
Joined: 25 Oct 2008
Posts: 610
Location: Kolkata,India
Followers: 9

Kudos [?]: 288 [0], given: 100

Re: probability:x,y,z [#permalink]  17 Aug 2009, 22:39
Quote:
you have answered to the question . "What is the probability of x OR y solving the problem?"

Z does not solve=x solves and Y,Z does not + y solves and x,z does not + x,y solve and z does not.Either of the three cases is possible...
_________________
Director
Joined: 01 Apr 2008
Posts: 906
Schools: IIM Lucknow (IPMX) - Class of 2014
Followers: 18

Kudos [?]: 302 [0], given: 18

Re: probability:x,y,z [#permalink]  18 Aug 2009, 01:01
P(x solving) = 1/4
P(y solving ) = 1/2
P(z not solving) = 1 - P (Z solving) = 3/8

Now we need a probability that x and y solves and z does not solve. AND means multiplication/ OR means addition.
Therefore, ans should be 3/64.
Re: probability:x,y,z   [#permalink] 18 Aug 2009, 01:01
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