- Suppose that a department contains 12 men and 17 women. How many ways are ther

Question

- Suppose that a department contains 12 men and 17 women. How many ways are there to form a committee with 6 members if it must have strictly more women than men?

-A bowl contains 7 red balls and 7 blue balls. A woman selects balls at random without looking at them.

(a) How many balls must she select (minimum) to be sure of having at least three blue balls?

(b) How many balls must she select (minimum) to be sure of having at least three balls of the same color?

-Consider the probablility model with sample space {A,B,C} and P(A)=0.1, P(B)=0.5, P(C)=0.4.Then

(a) P( B or C ) =

(b) P( A and B ) =

Explanation / Answer

You have to do this in "steps" or "levels". First, figure out how many ways the committee can be formed. Since the committee must have more women then men, you could have the following combinations of men and women...

6W, 0M
5W, 1M
4W, 2M

Now look at how many ways each committee can be formed. First look at the committee with 6 women and 0 men. Then we have...

17 choices for the first member of the committee
16 choices for the second member
15 for the 3rd
14 for the 4th
13 for the 5th
12 for the 6th

Thus, overall, there are 17*16*15*14*13*12 = 8910720 different committees.

Now look at the committee of 5 women and 1 man. Similarly, there are (17*16*15*14*13)*(12) = 8910720 different committees.

Lastly, for 4 women and 2 men, there are (17*16*15*14)*(12*11) = 7539840 possibilities.

Add these all together to get

8910720 + 8910720 + 7539840  = 25361280 total possible committees with more women than men

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