According to a recent student poll, 5/7 out of 21 members of

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BTGmoderatorLU: Moderator; Posts: 2505; Joined: Sun Oct 15, 2017 1:50 pm; Followed by:6 members

According to a recent student poll, 5/7 out of 21 members of

by BTGmoderatorLU » Fri Aug 09, 2019 5:18 pm

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Source: Veritas Prep

According to a recent student poll, $\frac{5}{7}$ 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. $\frac{1}{14}$
B. $\frac{4}{49}$
C. $\frac{2}{7}$
D. $\frac{45}{49}$
E. $\frac{13}{14}$

The OA is E

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Source: Beat The GMAT — Problem Solving | Fri Aug 09, 2019 5:18 pm

deloitte247: Legendary Member; Posts: 2214; Joined: Fri Mar 02, 2018 2:22 pm; Followed by:5 members

by deloitte247 » Sat Aug 10, 2019 12:31 pm

Given that 5/7 of 21 is interested in banking.
$$\frac{5}{7}\cdot21=15$$
So, 15/21 is interested in investment banking while 6 are NOT interested in investment banking.
But, we are asked to select 2 at random.
Using the logical inverse of at least one = none at all
Probability of "at least 1" want investment banking = PA
The probability that 2 do not want investment banking = P(Not A)
$$1-P\left(Not\ A\right)=PA$$
P(Not A) = Probability Inverse = (probability that 1 do not want banking) * (probability that 2 do not want banking)
$$=\frac{6}{21}\cdot\frac{5}{20}=\frac{30}{420}=\frac{3}{42}=\frac{1}{14}$$
$$PA=1-P\left(Not\ A\right)$$
$$PA=\frac{1}{1}-\frac{1}{14}=\frac{14-1}{14}=\frac{13}{14}$$

Answer = option E

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swerve: Legendary Member; Posts: 2499; Joined: Sun Oct 29, 2017 2:04 pm; Followed by:6 members

According to a recent student poll, 5/7 out of 21 members of

by swerve » Sun Aug 11, 2019 1:10 pm

15 students are interested, 6 are not interested

Prob $= 1 - \frac{6C_2}{21C_2} = 1 - \frac{6\cdot 5}{21\cdot 20}=1 - \frac{1}{14} = \frac{13}{14}\,\Rightarrow$ __E__

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Scott@TargetTestPrep: GMAT Instructor; Posts: 8086; Joined: Sat Apr 25, 2015 10:56 am; Location: Los Angeles, CA; Thanked: 43 times; Followed by:29 members

by Scott@TargetTestPrep » Wed Aug 14, 2019 7:03 pm

BTGmoderatorLU wrote:Source: Veritas Prep

According to a recent student poll, $\frac{5}{7}$ 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. $\frac{1}{14}$
B. $\frac{4}{49}$
C. $\frac{2}{7}$
D. $\frac{45}{49}$
E. $\frac{13}{14}$

The OA is E

We can use the equation 1 - P(none interested in banking) = P(at least one is interested in banking)

First, we see that 5/7 x 21 = 15 are interested in banking, so 6 aren't interested in banking and thus:

P(none interested in banking) = 6/21 x 5/20 = 2/7 x 1/4 = 2/28 = 1/14

Thus, P(at least one is interested in banking) = 1 - 1/14 = 13/14.

Answer: E

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