Question

In a class of 28 students, the teacher selects four people at random to participate in a geography contest. What is the probability that this group of four students includes at least two of the top three geography students in the class

Answer 1

Using the hypergeometric distribution, it is found that there is a 0.0452 = 4.52% probability that this group of four students includes at least two of the top three geography students in the class.

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

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.

In this problem:

• There are 28 students in class, hence $N = 28$.

• Four people will be chosen at random, hence $n = 4$.

• The top three is composed by 3 people, hence $k = 3$.

The probability that this group of four students includes at least two of the top three geography students in the class is:

$P(X \geq 2) = P(X = 2) + P(X = 3)$

In which:

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

$P(X = 2) = h(2,28,4,3) = \frac{C_{3,2}C_{25,2}}{C_{28,4}} = 0.0440$

$P(X = 3) = h(3,28,4,3) = \frac{C_{3,3}C_{25,1}}{C_{28,4}} = 0.0012$

Then:

$P(X \geq 2) = P(X = 2) + P(X = 3) = 0.0440 + 0.0012 = 0.0452$

0.0452 = 4.52% probability that this group of four students includes at least two of the top three geography students in the class.

You can learn more about the hypergeometric distribution at brainly.com/question/4818951