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 Post subject: Combinatorics and Probability (Vol. XII), Question #56.Posted: Sat Oct 30, 2010 8:50 pm

Joined: Thu Jul 01, 2010 12:05 am
Posts: 35
This question has the following situation: "The acceptance rate at a certain business school is 15% for first-time applicants and 20% for all re-applicants. If David is applying for admission for the first time this year, what is the probability that he will have to apply no more than twice before he is accepted?"

(A) 20%
(B) 30%
(C) 32%
(D) 35%
(E) 40%

The correct answer is (C). I viewed this question as a complimentary events equation where the probability of occurrence = 1 - probability of non-occurrence. The probability of non-occurrence (i.e. rejection) is 85% for first-time applicants and 80% for re-applicants. I subtracted and simplified the answer, and I arrived at 32%. Could you comment if my approach to this question is acceptable? Should I follow the solution instead? I am unsure about the solution because I cannot follow the two-scenario situation. Any recommendations?

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 Post subject: Re: Combinatorics and Probability (Vol. XII), Question #56.Posted: Mon Nov 08, 2010 12:30 pm

Joined: Thu Feb 12, 2009 6:32 pm
Posts: 497
There's not a good reason to go with a complementary events strategy on this one. In fact, it will probably make the problem more complicated, though you did arrive at the correct answer.
I'm not, however, sure how you approached it specifically, so I cant comment on whether it was correct or whether you got lucky...

There are two ways he can get admitted in the first 2 tries.
Either he gets admitted the first time, or he gets rejected the first time and admitted the second time.
The odds of the first scenario:
15%
The odds of the second scenario:
85% (rejection) * 20% = 17%
Add these together, since they're two different possibilities that fulfill our requirements, and we get 32%.

Veritas Help

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