Problem 21: Suppose there is a group of 10 boys and 12 girls. You choose a group

Question

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?

Explanation / Answer

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

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