total number of choices
= C(4,2) = 6
number of ways to pick 2 women
= C(2,2) = 1
number of ways with at least 1 man
= 1 - 1/6
= 5/6
Proof:
let the women be A, and B
let the men be C and D
choices
AB AC AD
BC BD
CD
of those 5 cases contain A or B
only one does not contain either an A or a B
My answer is correct
A committee of two persons is selected from two men and two women.
Find the probability that the committee will have at least one man.
1 answer
1 answer