C(4, 1)C(6, 1) is not the correct answer to “How many ways can a committee
of 2 be chosen from 4 men and 3 women and it must include at least 1 man?”
as the correct answer is C(4, 1)C(6, 1)–C(4, 2) because C(4, 2) is the number
of committees with two men and we are counting it twice if we simply use
C(4, 1)C(6, 1). Explain the situation. Why are we counting it twice?
Who do it in such a convoluting way, and why ask to explain a complicated
incorrect method??
all we have to do is exclude the case of 2 women
prob(exactly 2 women) = C(3,2) / C(7,2) = 3/21 = 1/7
so Prob(number of ways a committee of 2 can be chosen from 4 men and 3 women and it must include at least 1 man)
= 1 - 1/7 = 6/7
or
2 cases:
1 man and 1 woman, or 2 men
prob(of your stated event) = ( C(3,1)*C(4,1) + C(3,0)*C(4,2) ) / C(7,2)
= (3*4 + 1*6)/21
= 18/21 = 6/7