in a club with 9 male and 11 female members, how many 5 memeber committees can be chosen that have..

A.) at least 4 women?

B.)no more than 2 men?

Guest Nov 9, 2014

#1**+13 **

ncr(20,5)= 15504 ----Total number of ways to select 5 of twenty.

-------------------------

A.) Solution for at least 4 women

ncr(11,5)*ncr(9,0) = 462 Number of ways to select 5 of 11 females and select 0 of 9 males

ncr(11,4)*ncr(9,1) = 2970 Number of ways to select 4 of 11 females and select 1 of 9 males

Sum of combinations total = **3432**. So there are **3432** ways to select this group of 5 where 4 or more are women.

(3432/15504 = 22.1%) Incidental probability of this occurring.

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B.) Solution for no more than 2 men.

ncr(11,5)*ncr(9,0)= 462 Number of ways to select 5 of 11 females and select 0 of 9 males.

ncr(11,4)*ncr(9,1) = 2970 Number of ways to select 4 of 11 females and select 1 of 9 males.

ncr(11,3)*ncr(9,2) = 5940 Number of ways to select 3 of 11 females and select 2 of 9 males.

Sum of combinations total = **9372**. So there are **9372** ways to select this group of 5 where 2 or less are men.

(9372/15504 = 60.4%) Incidental probability of this occurring.

I doubled checked this, but Melody, CPhill, or Alan should verify the answers.

PS This assumes males are men and females are women. My dad says it is not easy to tell any more. So far, I don’t seem to have a problem. LOL

~7UP~

SevenUP
Nov 9, 2014

#1**+13 **

Best Answer

ncr(20,5)= 15504 ----Total number of ways to select 5 of twenty.

-------------------------

A.) Solution for at least 4 women

ncr(11,5)*ncr(9,0) = 462 Number of ways to select 5 of 11 females and select 0 of 9 males

ncr(11,4)*ncr(9,1) = 2970 Number of ways to select 4 of 11 females and select 1 of 9 males

Sum of combinations total = **3432**. So there are **3432** ways to select this group of 5 where 4 or more are women.

(3432/15504 = 22.1%) Incidental probability of this occurring.

-------------------------

B.) Solution for no more than 2 men.

ncr(11,5)*ncr(9,0)= 462 Number of ways to select 5 of 11 females and select 0 of 9 males.

ncr(11,4)*ncr(9,1) = 2970 Number of ways to select 4 of 11 females and select 1 of 9 males.

ncr(11,3)*ncr(9,2) = 5940 Number of ways to select 3 of 11 females and select 2 of 9 males.

Sum of combinations total = **9372**. So there are **9372** ways to select this group of 5 where 2 or less are men.

(9372/15504 = 60.4%) Incidental probability of this occurring.

I doubled checked this, but Melody, CPhill, or Alan should verify the answers.

PS This assumes males are men and females are women. My dad says it is not easy to tell any more. So far, I don’t seem to have a problem. LOL

~7UP~

SevenUP
Nov 9, 2014