Question

6 women and 7 men were randomly divided into two groups, with 9 individuals in one group and 4 individuals in the other group.

a) Calculate the probability that there will be at least two women in each of the groups.

b）Calculate the probability that there are at least five men in the 9-person group, if we know that there is at least one women in the 4-person group.

c) Calculate the expected value of the number of women in the 9-person group.

Answer 1

There are total 6+7=13 people. Number of ways of dividing them in two groups is

a)

Let G1 shows the 9 person groups and G2 shows the 4 person groups. Following table shows the number of women and men under the given condition " there will be at least two women in each of the groups" is

G1		G2
W	M	W	M
2	7	4	0
3	6	3	1
4	5	2	2

We need to take in to account the section of one group only, rest persons will automatically in second group. The total number of ways of above selection is;

The probability that there will be at least two women in each of the groups is

435 /715 = 0.6084

b)

Following table shows the possible number of men and women in both groups with the condition that "there is at least one women in the 4-person group";

G1		G2
W	M	W	M
5	4	1	3
4	5	2	2
3	6	3	1
2	7	4	0

We need to take in to account the section of one group only, rest persons will automatically in second group. The total number of ways of above selection is;

Out of above selections, following have at least five men in the 9-person group:

G1		G2
W	M	W	M
4	5	2	2
3	6	3	1
2	7	4	0

The total number of ways of above selection is;

The probability that there are at least five men in the 9-person group, if we know that there is at least one women in the 4-person group

470 / 680 = 0.6912

c)

Here we can use hyper-geometric distribution with parameters are follow:

Population size; N=13

Number of women in population: M = 6

Sample size: n= 9

The  expected value of the number of women in the 9-person group is