Question

There are 11 students on a committee. To decide which 4 of these students will attend a conference, 4 names are chosen at random by pulling names one at a time from a hat. What is the probability that Sarah, Jamal, Kate, and Mai are chosen in any order

Answer 1

Answer:

0.003 = 0.3% probability that Sarah, Jamal, Kate, and Mai are chosen in any order.

Step-by-step explanation:

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

Hypergeometric distribution:

The probability of x successes is given by the following formula:

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

In which:

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.

Combinations formula:

$C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

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

In this question:

11 students means that $N = 11$

4 are Sarah, Jamal, Kate, and Mai, so $k = 4$

4 are chosen, which means that $n = 4$

What is the probability that Sarah, Jamal, Kate, and Mai are chosen in any order?

This is P(X = 4). So

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

$P(X = 4) = h(4,11,4,4) = \frac{C_{4,4}*C_{7,0}}{C_{11,4}} = 0.003$

0.003 = 0.3% probability that Sarah, Jamal, Kate, and Mai are chosen in any order.