Question

A committee of 4 is being formed randomly from the employees at a school: 6 administrators, 38 teachers, and 5 staff. What is the probability that all 4 members are administrators? Express your answer as a fraction in lowest terms or a decimal rounded to the nearest millionth.

Answer 1

Answer:

0.00007 (to the nearest millionth)

Step-by-step explanation:

The total number of ways to select 4 members who are administrators is: $\binom{6}{4}$ and the total number of ways to select 4 members is: $\binom{49}{4}$.

Thus, we deduce that the probability of selecting 4 admins is: $\frac{\binom{6}{4}}{\binom{49}{4}} = 0.000070796\dots \approx 0.00007$

Answer 2

Using the hypergeometric distribution, it is found that there is a 0.000071 probability that all 4 members are administrators.

The employees are chosen without replacement, which means that the hypergeometric distribution is used to solve this question.

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:

• Total of 6 + 38 + 5 = 49 employees, thus $N = 49$.
• 6 administrators, thus $k = 6$.
• 4 are chosen, thus $n = 4$

The probability that all 4 members are administrators is P(X = 4), thus:

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

$P(X = 4) = h(4,49,4,6) = \frac{C_{6,4}C_{43,0}}{C_{49,4}} = 0.000071$

0.000071 probability that all 4 members are administrators.

A similar problem is given at https://brainly.com/question/24826394