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Sagot :
The probability that the 5-member committee contains all men is; P(all men) = 0.051
According to the question, the club is made up of 12 male and 8 female members.
- Total number of members = 12 + 8 = 20 members.
If the 5-member committee will contain all men; the probability is thus;
Since, the probability is without replacement, after each selection, the number of male members and the total number of members reduces by 1 each time.
Therefore, the probability is as follows;
- (12/20) × (11/19) × (10/18) × (9/17) × (8/16).
Ultimately, the algebraic sum of the expression above yields the probability that the 5-member committee contains all men.
P(all men) = 95040/1860480
P(all men) = 33/646.
P(all men) = 0.051
Read more:
https://brainly.com/question/13604758
Applying the hypergeometric distribution, it is found that there is a 0.0511 = 5.11% probability that the committee contains all men.
The members are chosen without replacement, which means that the hypergeometric distribution is used.
Hypergeometric distribution:
[tex]P(X = x) = h(x,N,n,k) = \frac{C_{k,x}*C_{N-k,n-x}}{C_{N,n}}[/tex]
[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]
The parameters are:
- x is the number of successes.
- N is the size of the population.
- n is the size of the sample.
- k is the total number of desired outcomes.
In this problem:
- 12 + 8 = 20 members, thus [tex]N = 20[/tex]
- 5 are chosen, thus [tex]n = 5[/tex]
- 12 are male, thus [tex]k = 12[/tex]
- The probability of an all-male committee is P(X = 5), then:
[tex]P(X = x) = h(x,N,n,k) = \frac{C_{k,x}*C_{N-k,n-x}}{C_{N,n}}[/tex]
[tex]P(X = 5) = h(5,20,5,12) = \frac{C_{12,5}*C_{8,0}}{C_{20,5}} = 0.0511[/tex]
0.0511 = 5.11% probability that the committee contains all men.
A similar problem is given at https://brainly.com/question/24271316
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