Question

Out of five men and five women, we form a committee consisting of four different people. Assuming that each committee of size four is equally likely, find the probabilities of the following events: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?

Answer 1

Answer:

1. 10/21

2. 11/42

3.  41/42

4. 2/15

Step-by-step explanation:

Probability defines the ratio of the number of favorable outcomes to the total number of outcomes.

Combination Operations

$5C2 = \frac{5! }{2! (5-2)!} = \frac{5*4*3! }{2! 3!} =\frac{5*4 }{2*1} =\frac{20}{2}= 10$

$10C4 = \frac{10! }{4! (10-4)!} = \frac{10*9*8*7*6! }{4! 6!} =\frac{10*9*8*7 }{4*3*2*1} =\frac{5040}{24}= 210$

$5C3 = \frac{5! }{3! (5-3)!} = \frac{5*4*3! }{3! 2!} =\frac{5*4 }{2*1} =\frac{20}{2}= 10$

$5C1 = \frac{5! }{1! (5-1)!} = \frac{5*4! }{1! 4!} =\frac{5 }{1} = 5$

$8C2 = \frac{8! }{2! (8-2)!} = \frac{8*7*6! }{2! 6!} =\frac{8*7 }{2*1} = 4*7= 28$

$5C4 = \frac{5! }{4! (5-4)!} = \frac{5*4! }{4! 1!} =\frac{5 }{1} = 5$

$5C0 = \frac{5! }{0! (5-0)!} = \frac{5! }{0! 5!} = 1$

1. Probability of forming a committee consisting of two men and two women

= $\frac{5C2 * 5C2}{10C4} =\frac{10*10}{210} =\frac{10}{21}$

2. Probability of forming a committee that has more women than men

(NOTE: Since the committee  desired has only 4 members, forming a committee that has more women than men can only be possible if we have  4 women, 0 men or 3 women 1 man)

= $\frac{5C3 * 5C1 + 5C4*5C0}{10C4} = \frac{10 * 5 + 5*1 }{210} = \frac{50 + 5}{210}=\frac{55}{210} = \frac{11}{42}$

3. Probability of forming a committee that has at least one man

(NOTE: at least one man implies the committee must not have all women members )

=$1 - \frac{5C4}{10C4} = 1 - \frac{5}{210} = 1 - \frac{1}{42} = \frac{41}{42}$

4. Probability of forming a committee with both Alice and Bob are members of the committee (NOTE: If Alice and Bob are compulsory member of the committee, there are only  two more people to join the committee from the remaining 8 people)

$=\frac{8C2}{10C4} =\frac{28}{210} =\frac{2}{15}$

it be like a coin toss ut a little more complicated. For every toss we have 4 outcomes from 210 possililities