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

Question

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

Bunuel · Accepted Answer

It should be: 
P(first interested in banking)*P(second is not)+P(first is not interested)*P(second is)+P(both are interested)

\frac{15}{21}*\frac{6}{20}+\frac{6}{21}*\frac{15}{20}+\frac{15}{21}*\frac{14}{20}=\frac{13}{14} .

maratikus · Answer

15 students are interested, 6 are not interested

Prob = 1 - 6C2/21C2 = 1 - (6*5/(21*20))=1 - 1/14 = 13/14 -> E

BrentGMATPrepNow · Answer

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

bmwhype2 · Answer

1-none
none = 2/7 * 1/4 = 1/14
1-1/14 = 1/13

jackychamp · Answer

Just a quick question, maratikus how do I know have to use combination in probability,

I mean my logic was,

15 interested, 6 not interested.

(1/15)(1/6) + (1/15)(1/6) + (2/15)

i.e. selected one from each group or select 2 from the interested group.
This will insure we have atleast one from the interested.

Where am I going wrong?

Thanks
Jack

lsuguy7 · Answer

I get the 1-none 
and the 2/7 but where does the 1/4 come from?

maratikus · Answer

6 students are not interested in investments, 21 total -> the first prob 6/21=2/7, 
5 students are not interested in investments, 20 total -> the second 5/20 = 1/4.

bmwhype2 · Answer

whoops. subtracted wrong... :-D

maratikus · Answer

the first person we choose is either interested 15/21 (works)
or not interested 6/21 -> then the second person we choose has to be interested 15/20

15/21 + (6/21)*(15/20) = 5/7 + (2/7)*(3/4)=5/7+ 6/28 = 10/14 + 3/14 = 13/14 -> E

is that what you were considering?

jackychamp · Answer

Thanks maratikus.

I was counting the probability of atleast one twice.

i.e.

prob of both + prob of one + prob of one which is wrong.

Thanks for the explanation.

-Jack

GMATBLACKBELT · Answer

Easy way: None are interested in I banking. 5/7*21= 15 --> 21-15 = 6

6/21*5/20 --> 2/7*1/4 --> 2/28--> 1/14 so simple 1-1/14 --> 13/14

E

jimmyjamesdonkey · Answer

Can one explain where I botched this....

I did Prob of one interested + Prob of both interested

(10/21)(11/20) + (10/21)(9/20) = 10/21
(Prob of interested)(Prob not interested) + (Prob interested)(Prob interested) = 10/21

Where did I screw this up?

maratikus · Answer

it's not 10 and 11 but 6 and 15

jimmyjamesdonkey · Answer

Still not coming out right....

Now getting

(15/21)(6/20) + (15/21)(14/20) = 60/84. Anyone explain this?

mneeti · Answer

Can anyone explain why and how this approach is wrong? Thanks!

draditya · Answer

probability that first person is not interested in investment banking = 6/21 
probability that second person is not interested in investment banking given that first also is not interested = 5/20

probability that at least one of them is interested in investment banking 
 = 1 - probability neither one of them interested in investment banking 
 = 1 - ((6/21) *(5/20)) 
 = 13/14

EMPOWERgmatRichC · Answer

Hi All,

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

AweG · Answer

OFFICIAL SOLUTION
Solution: E

“The probability of at least one” is very complicated to determine – you would need to count every time it happens and then divide it by the total number of possibilities. There is a much easier approach: calculating the inverse probability and then subtract it from one to get the actual probability. (Remember: PA = 1 – P(Not A).)  The logical inverse of “at least one” is “none at all.”

For purposes of this problem, “none at all” means the probability of not picking one of the students interested in investment banking.  \frac{5}{7}  of the 21 total members of the finance club (or 15 members) are interested in investment banking. Thus, the chance of not picking an investment banker the first time around is  \frac{6}{21}  (since 6 are not interested.) Once you have picked a non-investment banker out of the mix, there are five remaining non-investment bankers out of 20 total left. Therefore, the chance of picking another non-investment banker is  \frac{5}{20} . Since we need to pick twice and have both of our choices be non-investment bankers, we must multiply the probabilities. The total inverse probability would be:

P(inverse)=\frac{6}{21}∗\frac{5}{20}=\frac{2}{7}∗\frac{1}{4}=\frac{1}{14} 
Notice that this is an option (answer choice “A”). To finish off this problem, you need to transform the inverse probability into the actual, by subtracting the inverse probability from 1. Therefore,

P(actual)=1−\frac{1}{14}=\frac{14}{14}−14=\frac{13}{14}

The answer is “E”.

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