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An applicant has to go through 3 successive tests.

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Director
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An applicant has to go through 3 successive tests. [#permalink] New post 20 May 2006, 05:08
An applicant has to go through 3 successive tests. Probability of her passing the first test is P. Probability of passing successive tests are P or P/2 according to whether she passed the last test or not. She is selected if she passes as least 2 tests. In terms of P, what is the probability of her selection?

A. ((P^2)/2).(1-P)
B. 2P.(1-P^2)
C. 2.(1-P).P^2
D. 2.P^2 - P^3
E. 4.P^2 - P^4
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 [#permalink] New post 20 May 2006, 05:25
very good question . According to me the answer should be D.

she will pass in following scenario

PPP --->P*P*P
PPF --->P*P*(1-p)
PFP --->p*(1-p)*(p/2)
FPP --->(1-p)*(p/2)*P

adding all of the four you get (2*P^2) - P^3

I'm curious to know the OA .
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 [#permalink] New post 20 May 2006, 05:37
(D) as well :)
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Re: Probability - Tests [#permalink] New post 20 May 2006, 05:44
gmatmba wrote:
An applicant has to go through 3 successive tests. Probability of her passing the first test is P. Probability of passing successive tests are P or P/2 according to whether she passed the last test or not. She is selected if she passes as least 2 tests. In terms of P, what is the probability of her selection?

A. ((P^2)/2).(1-P)
B. 2P.(1-P^2)
C. 2.(1-P).P^2
D. 2.P^2 - P^3
E. 4.P^2 - P^4


go with D.

= p^3 + [p^2(1-p)] + [p(1-p)(p/2)] + [(1-p)(p/2)p]
= p^3 + p^2-p^3 + 2p(1-p)(p/2)
= p^2 + p^2(1-p)
= p^2 + p^2 - p^3
= 2 p^2 - p^3
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 [#permalink] New post 20 May 2006, 05:52
ipc302 wrote:
very good question . According to me the answer should be D.

she will pass in following scenario

PPP --->P*P*P
PPF --->P*P*(1-p)
PFP --->p*(1-p)*(p/2)
FPP --->(1-p)*(p/2)*P

adding all of the four you get (2*P^2) - P^3

I'm curious to know the OA .


wow same approach...
  [#permalink] 20 May 2006, 05:52
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An applicant has to go through 3 successive tests.

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