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According to a recent student poll, 5/7 out of 21 members of the finan
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22 Oct 2021, 06:59
22 Oct 2021, 06:59
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According to a recent student poll, 5/7 out of 21 members of the finance club are interested in a career in investment banking. If two students are chosen at random, what is the probability that at least one of them is interested in investment banking?
A. 1/14
B. 4/49
C. 2/7
D. 45/49
E. 13/14
A. 1/14
B. 4/49
C. 2/7
D. 45/49
E. 13/14
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Re: According to a recent student poll, 5/7 out of 21 members of the finan
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22 Oct 2021, 07:18
22 Oct 2021, 07:18
Carcass wrote:
According to a recent student poll, 5/7 out of 21 members of the finance club are interested in a career in investment banking. If two students are chosen at random, what is the probability that at least one of them is interested in investment banking?
A. 1/14
B. 4/49
C. 2/7
D. 45/49
E. 13/14
A. 1/14
B. 4/49
C. 2/7
D. 45/49
E. 13/14
There are several ways to solve this question.
First, 15 members are interested in investment banking (IB) and 6 are NOT interested in IB
We want P(have at least 1 interested in IB)
When it comes to probability questions involving "at least," it's best to try using the complement.
That is, P(Event A happening) = 1 - P(Event A not happening)
So, here we get: P(have at least 1 interested in IB) = 1 - P(not have at least 1 interested in IB)
What does it mean to not have at least 1 interested in IB? It means getting ZERO members interested in IB.
So, we can write: P(have at least 1 interested in IB) = 1 - P(have ZERO interested in IB)
Okay, let's go...
P(have ZERO interested in IB) = P(1st person is NOT interested in IB AND 2nd person is NOT interested in IB)
= P(1st person is NOT interested in IB) x P(2nd person is NOT interested in IB)
= 6/21 x 5/20
= 1/14
So, P(at least 1 interested in IB) = 1 - P(not at least 1 interested in IB)
= 1 - 1/14
= 13/14
Answer: E
Cheers,
Brent
gmatclubot
Re: According to a recent student poll, 5/7 out of 21 members of the finan [#permalink]
22 Oct 2021, 07:18
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