Hi gmatdriller,

When it comes to probability questions, there are 2 things that you can calculate: what you WANT and what you DON'T WANT.

In probability, (the probability of what you WANT) + (the probability of what you DON'T WANT) = 1

In this question, we WANT AT LEAST 1 member chosen to be interested in investment banking; what we DON'T WANT is 0 members chosen to be interested in investment banking. The second option will be easier to calculate. Here's how:

We're told that 5/7 of the 21 members are interested in investment banking:

15 interested in investment banking
6 NOT interested in investment banking

We're asked to select 2 at random. Based on the above probability concepts....

1 - (probability that the 2 DON'T WANT investment banking) = the probability of AT LEAST 1 that does want investment banking

The probability that the 1st DOESN'T WANT investment banking = 6/21
The probability that the 2nd DOESN'T WANT investment banking = 5/20

(6/21)(5/20) = 30/420 = 3/42 = 1/14

1 - 1/14 = 13/14 = the probability that AT LEAST 1 of the 2 chosen is interested in investment banking.

Final Answer: E

GMAT assassins aren't born, they're made,
Rich

gmatdriller wrote:According to a recent student poll, 5/7 of the 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

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
= [spoiler]13/14[/spoiler]
= E

Cheers,
Brent

According to a recent student poll, 5/7 of the 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

Here's another (longer) approach:

We know that 15 members are INTERESTED
And 6 members are NOT interested.

We want P(have at least 1 INTERESTED)

There are THREE DIFFERENT CASES that satisfy this condition:
case 1: 1st person is INTERESTED and the 2nd person is NOT interested
The probability that this case occurs = (15/21)(6/20) = 3/14
case 2: 1st person is NOT interested and the 2nd person is INTERESTED
The probability that this case occurs = (6/21)(15/20) = 3/14
case 3: 1st person is INTERESTED and the 2nd person is INTERESTED
The probability that this case occurs = (15/21)(14/20) = 1/2

So, P(have at least 1 INTERESTED) = P(case 1 OR case 2 OR case 3)
= P(case 1) + P(case 2) + P(case 3)
= 3/14 + 3/14 + 1/2
= [spoiler]13/14[/spoiler]
= E

Cheers,
Brent

According to a recent student poll, 5/7 of the 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

We can also use COUNTING TECHNIQUES to answer the question.

P(have at least 1 interested in IB) = 1 - P(have ZERO interested in IB)

P(have ZERO interested in IB)
P(have ZERO interested in IB) = (# of ways to select 2 people that are NOT interested in IB)/(# of ways to select 2 people from 21 members)

# of ways to select 2 people that are NOT interested in IB: 6 members are NOT interested in IB
We can select 2 of these members in 6C2 ways.
6C2 = 15

# of ways to select 2 people from 21 members:
We can select 2 of the 21 members in 21C2 ways.
21C2 = 210

So, P(have ZERO interested in IB) = 15/210 = 1/14

So, P(have at least 1 interested in IB) = 1 - P(have ZERO interested in IB)
= 1 - 1/14
= [spoiler]13/14[/spoiler]
= E

If anyone is interested, we have a free video on calculating combinations (like 6C2 and 21C2) in your head: https://www.gmatprepnow.com/module/gmat-counting?id=789

Cheers,
Brent