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Probability help plz!

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Probability help plz!

by Tanjello » Fri May 21, 2010 4:58 pm
A club has exactly 3 men & 7 women members. If 2 members are randomly selected to be pres and VP respectively, and no member can hold 2 offices simultaneously, what's the probability that a woman is selected for at least one position?

A) 14/15
B) 4/5
C) 8/15
D) 7/15
E) 1/5

I get D if I use the permutation equation, but I get A if I find the P(all men), can someone please tell me which is the correct answer and why my other approach is incorrect? Thanks! :)

A) P(all men): 3/10*2/9=1/15 => P(at least 1 woman)= 1-1/15= 14/15
D) Perm(7,2)/Perm(10,2)= (7!/5!)(10!8!)=7/15
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by liferocks » Fri May 21, 2010 5:33 pm
Tanjello wrote: D) Perm(7,2)/Perm(10,2)= (7!/5!)(10!8!)=7/15
here you have considered the probability of selecting all women.
the probability that a woman is selected for at least one position=probability of selecting all women+probability of selecting only one women
=7C2/10C2 +7C1*3C1/10C2
=7*6/10*9+7*3*2/10*9
=7/15+7/15
=14/15
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by Thouraya » Wed Jun 09, 2010 5:40 am
Isn't it easier to consider the probability of not selecting women at all in this case? ie: 1-P(only men)?

Total probability= 10C2= 45

But what's the probability of men? 3C2?


Thank you
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by tpr-becky » Wed Jun 09, 2010 6:00 am
"At least one "probability is calculated by the formula 1-(probability of none)
Probability of getting none is done by creating a formula for each pick (and not getting a female) and then multiplying them.


In this case it would be 1 - (3/10 * 2/9) which is 1- (2/30) = which is 28/30 reduced to 14/15.
Last edited by tpr-becky on Sat Jun 12, 2010 12:31 pm, edited 1 time in total.
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by Thouraya » Sat Jun 12, 2010 11:10 am
U meant 14/15 Becky, Thank you!

Yup I got that, but why did we not divide further by the total number of possibilities (10C2=45); ie: 14/15/45?


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by Stuart@KaplanGMAT » Sat Jun 12, 2010 12:25 pm
Thouraya wrote:U meant 14/15 Becky, Thank you!

Yup I got that, but why did we not divide further by the total number of possibilities (10C2=45); ie: 14/15/45?


Thanks!
By using the probability formula, we've already divided through by the total number of possibilities, so there's no need to duplicate.

Your original post presents another method of attack to count the number of groups that include all men.

Prob (all men) = (# of selections of just men)/(total # of selections)

= 3C2/10C2

= 3/45

= 1/15

So Prob (not all men) = 1 - 1/15 = 14/15
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by Thouraya » Sun Jun 13, 2010 9:32 am
Thank you Stuart!

So from your reply, I understand that when I use the combinations formula, I shall divide by the total number; however, when I use the probabilities formulas (1-p(x) ) or x/total x y/total-1 I shouldnt divide by the total?

Another question on this problem if you dont mind..I looked at the post above but didn't really get it cuz if I want to solve this problem not from the combinations approach, but the usual probability:

We now know there are two methods: Either 1- P(choosing only men) = 1-(3/10 x 2/9)= 1-1/15 =14/15

OR

probability of choosing 2 women + probability of choosing 1 women and 1 men
= 7/10 x 6/9 + 7/10 x 3/9
=14/30 +
=7/15 +

I am aware that I am making a mistake in probability of choosing 1 men, but what exactly is the mistake in 3/9?

THANK YOU!
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