+556 The Grammar club has 20 members: 10 boys and 10 girls. A 4-person committee is chosen at random. What is the probability that the committee has at least 1 boy and at least 1 girl?
+33616 You are right - my answer above is incorrect!
Further thought suggests that it should be:
p(at least 1 boy and 1 girl) = 1 - (number of ways of choosing 4 boys from 10 + number of ways of choosing 4 girls from 10)/number of ways of choosing 4 people from 20)
or:
p(at least 1 boy and 1 girl) = 1 - 2*nCr(10,4)/nCr(20,4) → 295/323 → 0.913
+33616 p(at least 1 boy and at least 1 girl) = 1 - p(no boys) - p(no girls)
→ 1 - 2*ncr(20,4)(1/2)20
+33616 You are right - my answer above is incorrect!
Further thought suggests that it should be:
p(at least 1 boy and 1 girl) = 1 - (number of ways of choosing 4 boys from 10 + number of ways of choosing 4 girls from 10)/number of ways of choosing 4 people from 20)
or:
p(at least 1 boy and 1 girl) = 1 - 2*nCr(10,4)/nCr(20,4) → 295/323 → 0.913
