The probabilities of the following events are 10/21, 11/42,41/42, and 2/15 respectively
Probability of forming a committee consisting of two men and two women = (5C2 × 5C2) / 10C4
5C2 = 5! / 2!(5 - 2)!
= 5! / 2!3!
= 5×4× / 2×1
= 20/2
= 10
So,
(5C2 × 5C2) / 10C4
= 10×10 / 210
= 100/210
= 10/21
Probability of forming a committee consisting of two men and two women = 10/21
Probability of forming a committee that has more women than men. This means forming a committee that has 4 women, 0 men or 3 women 1 man
= {(5C3 × 5C1) + (5C4 × 5C0)} / 10C4
= {(10×5) + (5×1)} / 210
= (50 + 5) / 210
= 55/210
= 11/42
Probability of forming a committee that has at least one man. This means the committee must not have all women as members
= 1 - 5C4/10C4
= 1 - 5/210
= 1 - 1/42
= 41/42
Probability of forming a committee with both Alice and Bob are members of the committee. This means there are only two more people to join the committee from the remaining 8 people.
= 8C2 / 10C4
= 28/210
= 2/15
The complete question:
1. The committee consists of two men and two women.
2. The committee has more women than men.
3. The committee has at least one man. For the remainder of the problem, assume that Alice and Bob are among the ten people being considered.
4. Both Alice and Bob are members of the committee.Shouldn't it be like a coin toss but with more than just heads and tails?
Learn more about probability:
https://brainly.com/question/25870256
