Respuesta :
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
[tex]5C2 = \frac{5! }{2! (5-2)!} = \frac{5*4*3! }{2! 3!} =\frac{5*4 }{2*1} =\frac{20}{2}= 10[/tex]
[tex]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[/tex]
[tex]5C3 = \frac{5! }{3! (5-3)!} = \frac{5*4*3! }{3! 2!} =\frac{5*4 }{2*1} =\frac{20}{2}= 10[/tex]
[tex]5C1 = \frac{5! }{1! (5-1)!} = \frac{5*4! }{1! 4!} =\frac{5 }{1} = 5[/tex]
[tex]8C2 = \frac{8! }{2! (8-2)!} = \frac{8*7*6! }{2! 6!} =\frac{8*7 }{2*1} = 4*7= 28[/tex]
[tex]5C4 = \frac{5! }{4! (5-4)!} = \frac{5*4! }{4! 1!} =\frac{5 }{1} = 5[/tex]
[tex]5C0 = \frac{5! }{0! (5-0)!} = \frac{5! }{0! 5!} = 1[/tex]
1. Probability of forming a committee consisting of two men and two women
= [tex]\frac{5C2 * 5C2}{10C4} =\frac{10*10}{210} =\frac{10}{21}[/tex]
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)
= [tex]\frac{5C3 * 5C1 + 5C4*5C0}{10C4} = \frac{10 * 5 + 5*1 }{210} = \frac{50 + 5}{210}=\frac{55}{210} = \frac{11}{42}[/tex]
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 )
=[tex]1 - \frac{5C4}{10C4} = 1 - \frac{5}{210} = 1 - \frac{1}{42} = \frac{41}{42}[/tex]
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)
[tex]=\frac{8C2}{10C4} =\frac{28}{210} =\frac{2}{15}[/tex]
it be like a coin toss ut a little more complicated. For every toss we have 4 outcomes from 210 possililities
The number of ways a committee can be formed with Both Alice and Bob as members of the committee is 0.1334.
What is the Probability?
The probability helps us to know the chances of an event occurring.
[tex]\rm{Probability=\dfrac{Desired\ Outcomes}{Total\ Number\ of\ outcomes\ possible}[/tex]
We know that a committee is needed to be made out of 10 people consisting of five men and five women, also, the size of the committee is 4. therefore, the total number of possible ways in which a committee can be formed,
[tex]\begin{aligned}\text{ Total Number of ways a committee can be formed} &= \ ^{10}C_4\\&=\dfrac{10!}{4!(10-4)!}\\& = 210\\\end{aligned}[/tex]
1.) The committee consists of two men and two women.
In order to find the probability of a committee consisting of two men and two women we need to know the desired outcome, therefore, the number of ways in which a committee of 2 men and 2 women can be made, and the total possible outcome, therefore, the total number of possible ways in which a committee can be formed,
[tex]\begin{aligned}\text{Number of ways a committee of 2 men and 2 women} &= \ ^{5}C_2 \times ^{5}C_2\\&=100\end{aligned}[/tex]
Probability of forming a committee with 2 men and 2 women
[tex]\rm{Probability=\dfrac{Desired\ Outcomes}{Total\ Number\ of\ outcomes\ possible}[/tex]
[tex]\rm{Probability=\dfrac{100}{210} = 0.476[/tex]
2.) The committee had more women than men therefore, the committee may have either 3 women or 4 women,
[tex]\begin{aligned}\text{Number of ways a committee of 1 men and 3 women} &= \ ^{5}C_1 \times ^{5}C_3\\&=50\end{aligned}[/tex]
[tex]\begin{aligned}\text{Number of ways a committee of 4 women} &= \ ^{5}C_4\\&=5\end{aligned}[/tex]
[tex]\rm{Probability=\dfrac{Desired\ Outcomes}{Total\ Number\ of\ outcomes\ possible}[/tex]
[tex]\rm{Probability=\dfrac{50+5}{210}=0.2619[/tex]
3.) The committee has at least one man
We need to find the probability that the committee has at least one man, although it can have more than one man, therefore,
[tex]\begin{aligned}\text{Number of ways a committee of 1 men and 3 women} &= \ ^{5}C_1 \times ^{5}C_3\\&=50\end{aligned}[/tex]
[tex]\begin{aligned}\text{Number of ways a committee of 2 men and 2 women} &= \ ^{5}C_2 \times ^{5}C_2\\&=100\end{aligned}[/tex]
[tex]\begin{aligned}\text{Number of ways a committee of 3 men and 1 women} &= \ ^{5}C_3 \times ^{5}C_1\\&=50\end{aligned}[/tex]
[tex]\begin{aligned}\text{Number of ways a committee of 4 men} &= \ ^{5}C_4\\&=5\end{aligned}[/tex]
[tex]\rm{Probability=\dfrac{Desired\ Outcomes}{Total\ Number\ of\ outcomes\ possible}[/tex]
[tex]\rm{Probability=\dfrac{50+100+50+5}{210} = 0.97619[/tex]
4.) Both Alice and Bob are members of the committee.
Since we need to find the number of ways a committee can be formed with Both Alice and Bob as members of the committee.
[tex]\text{Number of ways other 2 people can be selected} = ^8C2 = 28[/tex]
[tex]\rm{Probability=\dfrac{Desired\ Outcomes}{Total\ Number\ of\ outcomes\ possible}[/tex]
[tex]\rm{Probability=\dfrac{28}{210} = 0.1334[/tex]
Hence, the number of ways a committee can be formed with Both Alice and Bob as members of the committee is 0.1334.
Learn more about Probability:
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