Question

4-member curling team is randomly chosen from 6 grade 11 students and 8 grade 12 students. What is the probability that the team has at least 2 grade 11 students?

Answer 1

Answer:

0.5944 = 59.44% probability that the team has at least 2 grade 11 students

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.

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

We have that:

6 + 8 = 14 students, which means that $N = 14$

6 grade 11 students means that $k = 6$

Teams of 4 members means that $n = 4$

What is the probability that the team has at least 2 grade 11 students?

This is:

$P(X \geq 2) = 1 - P(X < 2)$

In which

$P(X < 2) = P(X = 0) + P(X = 1)$

So

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

$P(X = 0) = h(0,14,4,6) = \frac{C_{6,0}*C_{8,4}}{C_{14,4}} = 0.0699$

$P(X = 1) = h(1,14,4,6) = \frac{C_{6,1}*C_{8,3}}{C_{14,4}} = 0.3357$

Then

$P(X < 2) = P(X = 0) + P(X = 1) = 0.0699 + 0.3357 = 0.4056$

$P(X \geq 2) = 1 - P(X < 2) = 1 - 0.4056 = 0.5944$

0.5944 = 59.44% probability that the team has at least 2 grade 11 students