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# An applicant has to go through 3 successive tests.

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Director
Joined: 16 Aug 2005
Posts: 937

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Location: France
An applicant has to go through 3 successive tests. [#permalink]

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20 May 2006, 06:08
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

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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Manager
Joined: 13 Dec 2005
Posts: 224

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Location: Milwaukee,WI

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20 May 2006, 06:25
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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SVP
Joined: 01 May 2006
Posts: 1794

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20 May 2006, 06:37
(D) as well

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VP
Joined: 29 Dec 2005
Posts: 1338

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20 May 2006, 06:44
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

Kudos [?]: 69 [0], given: 0

VP
Joined: 29 Dec 2005
Posts: 1338

Kudos [?]: 69 [0], given: 0

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20 May 2006, 06:52
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...

Kudos [?]: 69 [0], given: 0

20 May 2006, 06:52
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