Probability Question (2)?

There are 11 C 3 ways to choose the committee, or

(11!)/(8!*3!) = 165

There are 5 ways to select the one man for the desired committee, 1 way to select Jane, and 5 ways to select the other woman.

5*1*5 = 25 different committees which meet the criteria.

So the probability of choosing that particular committee, assuming it's random, is

25/165 = 5/33

To calculate the probability, you need to count the number of desirable outcomes (i.e. the number of ways to select a 1 man, 1 woman, and 1 Jane committee) and divide that by the total number of outcomes (i.e. the number of 3 person committees that can be made from Jane and 2 other people).

First we'll count the number of committees that are desirable. We need one man (we have 5 choices), one Jane (one choice), and one other woman (5 choices). So the total number will be:

5 * 1 * 5 = 25

Now we count the total number of possible committees. We have one choice for the Jane, and 10 other people to choose 2 people from. The number will be:

1 * (10 C 2)

Where 10 C 2 is the number of unordered selections of 2 objects from 10 objects. Many scientific calculators can calculate selections like this, and it comes to 45. So the probability will be:

P = 25/45 = 5/9

Note: The degree of the quotient is one less than the degree of the dividend. And the degree of the remainder is less than the degree of the divisor, x + 3, which in this case is 1. The remainder therefore is of degree 0, which is a number.

In general, if we divide a polynomial of degree n by a polynomial of degree 1, then the degree of the quotient will be n − 1. And the remainder will be a number.

P(Jane, a woman and a man)

= 1C1 * 5C1 * 5C1 / 11C3

= 1*5*5 / 165

= 5 / 33