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A team of 8 people exists, and they plan to form a committe [#permalink]
09 Jun 2004, 19:48

A team of 8 people exists, and they plan to form a committe of 3. Given that both Noel and Asshie are on the team, what is the probability that Noel is on the committe but NOT Ashie?

Got 9/56
# of ways to select a committee of 3 out of 8 people: 8C3 = 56

When Noel is on committee, total # of ways to select other 2 people: 7C2 = 21

(unfavorable outcome) # of ways to select remaining 1 person when Ashie is in: 6C1 = 6 --> this can be interchanged 2 ways for Ashie can be selected first or second out of the two --> 6C1 * 2 = 12

Answer should be 21-12 / 56 = 9/56 _________________

Re: conditional probability [#permalink]
10 Jun 2004, 03:58

lastochka wrote:

A team of 8 people exists, and they plan to form a committe of 3. Given that both Noel and Asshie are on the team, what is the probability that Noel is on the committe but NOT Ashie?

It is C[2,6]/C[3,8] = [6!/(4!2!)]/[8!/(5!3!)] = 15/56.

I kind of doubted my answer too. When I said 6C1 * 2, the answer should have been 6C1 and we do not need to multiply by 2 since there is only 1 spot to take when Ashie is out. Hence, answer should have been 21-6 / 56 = 15/56
But your solution manan is much more simple. Neat _________________

Re: conditional probability [#permalink]
10 Jun 2004, 04:37

lastochka wrote:

A team of 8 people exists, and they plan to form a committe of 3. Given that both Noel and Asshie are on the team, what is the probability that Noel is on the committe but NOT Ashie?

# of ways to select committe of 3 out of 8 people - 8C3 = 56
Ashie is not there in 56(total selections) - 7C2(ashie will be there)
= 56 - 21 = 35 selections
# of selections in which both are not there: 6C3 = 20
So to find # of selection Ashie wont be there but Noel will be there
= 35 - 20 = 15

So, Probability that Noel is on the committe but NOT Ashie = 15 / 56

side note: calculating completely in each step may not be good idea (like 8C3 , just keep it like that,donot calculate it).

in the last step you get 6C2/*8C3 now solving will be easy as you will find few common ones in the numerator and denominator.

This tip may not be that useful here as we are only dealing with 3 items here ,but if the numbers are big like 15C6/17C7 there is no point in calcuating 15C6 and 17C7 seperately, unncessarily kills time. _________________

side note: calculating completely in each step may not be good idea (like 8C3 , just keep it like that,donot calculate it).

in the last step you get 6C2/*8C3 now solving will be easy as you will find few common ones in the numerator and denominator.

This tip may not be that useful here as we are only dealing with 3 items here ,but if the numbers are big like 15C6/17C7 there is no point in calcuating 15C6 and 17C7 seperately, unncessarily kills time.

Good point GoalStanford, but what if you have only say 1/17C7 you can calculate it 17!/(10!*7!) or write immediately 17*16*15*14*13*12*11/7!. In any case everyone will uses his own approach. I for example tend to use your approach, GoalStanford, but one my friend uses the approach i described. _________________

Re: conditional probability [#permalink]
11 Jun 2004, 04:33

lastochka wrote:

A team of 8 people exists, and they plan to form a committe of 3. Given that both Noel and Asshie are on the team, what is the probability that Noel is on the committe but NOT Ashie?

I forgot the solution. Assuming that Noel is on the committee we have two vacan seats and choose out of 7-Ashie=6 people.

Good point GoalStanford, but what if you have only say 1/17C7 you can calculate it 17!/(10!*7!) or write immediately 17*16*15*14*13*12*11/7!. In any case everyone will uses his own approach.

I would use 17C7 till the last step....yeah, i agree it is upto one's comfortable level also. _________________