Question about how to solve combination problem

[OtZman]

I'm taking a statistics course right now, so chances are I'll ask questions about this subject again the near future.

This is a simple problem, but I'm having trouble solving it:

"A committee of 7 members is to be formed by selecting members at random from a pool of 14 candidates consisting of 5 women and 9 men.

What is the probability that there will be at least three women in the committee?"

This is how I tried to solve it:

7 members are to be picked, and we are supposed to figure out the probability of 3 or more of them being women. This is the same as 1 - [the probability that number of women < 3]:

p = probability of picking a woman = 5/14

1 - ( P(X = 0) + P(X = 1) + P(X = 2) ) = 1 - [ 7!(0!7!) * p^0 * (1-p)^7 + 7!/(1!6!) * p^1 * (1-p)^6 + 7!/(2!5!) * p^2 * (1-p)^5 ] = 0.484087... (if I've calculated correctly)

This was however incorrect according to the book. If someone would please let me know how to correctly calculate this it'd be great! Also, be as thorough as possible. /images/graemlins/smile.gif

[PairTheBoard]

[ QUOTE ]
p = probability of picking a woman = 5/14


[/ QUOTE ]

When? On the first pick? After a man has been chosen? After a woman has been chosen?

Think of the Number of Ways you can choose 7 from 14. Then think of how many of these ways have 0 women. How many have exactly 1 woman? How many have exactly 2 women? It's a combinatorics problem. You are not applying the tools of combinarics.

PairTheBoard

[BruceZ]

[ QUOTE ]
I'm taking a statistics course right now, so chances are I'll ask questions about this subject again the near future.

This is a simple problem, but I'm having trouble solving it:

"A committee of 7 members is to be formed by selecting members at random from a pool of 14 candidates consisting of 5 women and 9 men.

What is the probability that there will be at least three women in the committee?"

This is how I tried to solve it:

7 members are to be picked, and we are supposed to figure out the probability of 3 or more of them being women. This is the same as 1 - [the probability that number of women < 3]:

p = probability of picking a woman = 5/14

1 - ( P(X = 0) + P(X = 1) + P(X = 2) ) = 1 - [ 7!(0!7!) * p^0 * (1-p)^7 + 7!/(1!6!) * p^1 * (1-p)^6 + 7!/(2!5!) * p^2 * (1-p)^5 ] = 0.484087... (if I've calculated correctly)

This was however incorrect according to the book. If someone would please let me know how to correctly calculate this it'd be great! Also, be as thorough as possible. /images/graemlins/smile.gif

[/ QUOTE ]

First of all, this belongs on the probability forum.

Second, you don't want the binomial distribution as that is only for independent trials. These trials aren't independent since each time you select a person the probability for the next person changes. You want the hypergeometric distribution.

P(3 or more women) = 1 - P(0 women) - P(1 woman) - P(2 women)

= 1 -
C(5,0)*C(9,7)/C(14,7) -
C(5,1)*C(9,6)/C(14,7) -
C(5,2)*C(9,5)/C(14,7)

= 50%.

[GMontag]

There are 14c7=3432 possible selections of committe members.

For 0 women, there are 9c7=36 possible selections.
For 1 woman, there are 5*9c6=420 possible selections.
For 2 women, there are 5c2*9c5=1260 possible selections.

The number of possible selections with 3 or more women is therefore 3432-(1260+420+36)=1716, and therefore the probability of picking at least 3 women is 1716/3432=.5