From a group of 3 boys and 3 girls, 4 children are to be randomly selected. What is the probability that equal numbers of boys and girls will be selected?
Prof what about this one 3C2x3C2/6C4?
But why should we multiply here?
3 boys = a, b and c
3 girls = x, y and z
no of ways 2 boys can be chosen = ab, bc, and ca
no of ways 2 girls can be chosen = xy, yz, and zx
no of ways 2boys and 2 girls can be chosen = (ab+xy), (ab+yz), ........................ and (ca+zx) = 9 ways
no of ways 4 out of 6 can be chosen = 6c4 = 15
so the prob = 9/15 = 3/5