Problem 21 Suppose there is a group of 10 boys and 12 girls

Problem 21: Suppose there is a group of 10 boys and 12 girls. You choose a group of people using a random procedure as follows. For every person you toss a fair coin - if it turns up head you include that person, otherwise you do not include him or her.

1. What is the probability that you choose a group of size 6?

2. What is the probability that you choose 5 boys and 6 girls in the group?

3. What is the expected number of people in the group?

4. What is the expected number of boy,girl pairs in the group?

Solution

Total number of ways of forming a group is 2^(22)=22C0+22C1+....+22C22

1)probability of choosing group of size 6 is = 22C6/2^(22)

2)total number of ways of selecting 11 is 22C11

can be (10b,1g),(9b,2g),(8b,3g),(7b,4g),(6b,5g),(5b,6g),....(0b,11g)

P(5b,6g)=10C5*12C6

hence 10C5*12C6/2^(22)

3)

E(X)=i(22Ci) where i ranges from 0 to22

4)E(X)=0*(10C0*2^12+2^10*12C0)+

1*(10C1*2^11+12C1*2^9)+...

=i*(10Ci*2^(12-i)+2^(10-i)*12Ci) where i ranges from 0 to 10