Question

A group of 22 people is divided into two groups of 11: Group A and Group B (note that the labels distinguish the groups and you do not have to worry about double counting). What is the probability that two friends, Jim and Tom, are not in the same group?

Answer 1

Solution:

Two group of 10 out of 20 can be make in ways

Total required probability for being in different group is the sum of probability A and B

I.e. P = P(A) + P(B)

A: let A be the probability that Jim are in the group A

So probability that Jim are in group A is 11 out of 22 and tom in group B is 11out of 21

P(A) = (11/22 × 11/21)

B : Let B be the probability that Tom are in group A and Jim are in group B

P (B)= (11/22 ×11/21)

Required probability = P(A) +P(B) = 0.5238