Question

Membership selection. A town council has ​members, Democrats and Republicans.

​(A)

If the president and​ vice-president are selected at​ random, what is the probability that they are both​ Democrats?

​(B)

If a​ 3-person committee is selected at​ random, what is the probability that Republicans make up the​ majority?

The probability that they are both Democrats is approximately

enter your response here.
​(Type a decimal. Round to three decimal​ places.)

Answer 1

Using the hypergeometric distribution, the probabilities are given as follows:

a) 0.2 = 20%.

b) 0.5539 = 55.39%.

What is the hypergeometric distribution formula?

The formula is:

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

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

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.

Researching this problem on the internet, it is found that the council has 15 members, of which 7 are Democrats and 8 are Republicans.

Item a:

The parameters are:

N = 15, n = 2, k = 7.

The probability is P(X = 2), hence:

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

$P(X = 2) = h(2,15,2,7) = \frac{C_{7,2}C_{8,0}}{C_{15,2}} = 0.2$

Item b:

At most one Democrat, with n = 3, hence:

$P(X \leq 1) = P(X = 0) + P(X = 1)$

In which:

$P(X = 0) = h(0,15,3,7) = \frac{C_{3,0}C_{8,3}}{C_{15,3}} = 0.1231$

$P(X = 1) = h(1,15,3,7) = \frac{C_{3,1}C_{8,2}}{C_{15,3}} = 0.4308$

Then:

$P(X \leq 1) = P(X = 0) + P(X = 1) = 0.1231 + 0.4308 = 0.5539$

More can be learned about the hypergeometric distribution at https://brainly.com/question/24826394

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