Question

In a club with 12 male and 8 female​ members, a 5​-member committee will be randomly chosen. Find the probability that the committee contains all men.

Answer 1

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:

$P(X = x) = h(x,N,n,k) = \frac{C_{k,x}*C_{N-k,n-x}}{C_{N,n}}$

$C_{n,x} = \frac{n!}{x!(n-x)!}$

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 $N = 20$
• 5 are chosen, thus $n = 5$
• 12 are male, thus $k = 12$
• The probability of an all-male committee is P(X = 5), then:

$P(X = x) = h(x,N,n,k) = \frac{C_{k,x}*C_{N-k,n-x}}{C_{N,n}}$

$P(X = 5) = h(5,20,5,12) = \frac{C_{12,5}*C_{8,0}}{C_{20,5}} = 0.0511$

0.0511 = 5.11% probability that the committee contains all men.

A similar problem is given at brainly.com/question/24271316

Answer 2

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:

brainly.com/question/13604758