Question

In a MBS first year class, there are three sections each including 20 students. In the first section, there are 10 boys and 10 girls, in the second section there are 15 boys and 5 girls, and in the third section, there are 12 boys and 8 girls. Five students are selected at random from each group to form committee of 15 students. What is the probability that all the 15 students selected are girls?​

Answer 1

Answer:

$3.52 \times 10^{-9} = 3.52 \times 10^{-7}\%$ probability that all the 15 students selected are girls

Step-by-step explanation:

The selection is from a sample without replacement, so we use the hypergeometric distribution to solve this question.

Hypergeometric distribution:

The probability of x sucesses 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 sucesses.

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.

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

All girls from the first group:

20 students, so $N = 20$

10 girls, so $k = 10$

5 students will be selected, so $n = 5$

We want all of them to be girls, so we find P(X = 5).

$P_1 = P(X = 5) = h(5,20,5,10) = \frac{C_{10,5}*C_{10,5}}{C_{20,5}} = 0.0163$

All girls from the second group:

20 students, so $N = 20$

5 girls, so $k = 5$

5 students will be selected, so $n = 5$

We want all of them to be girls, so we find P(X = 5).

$P_2 = P(X = 5) = h(5,20,5,5) = \frac{C_{5,5}*C_{15,5}}{C_{20,5}} = 0.00006$

All girls from the third group:

20 students, so $N = 20$

8 girls, so $k = 8$

5 students will be selected, so $n = 5$

We want all of them to be girls, so we find P(X = 5).

$P_3 = P(X = 5) = h(5,20,5,8) = \frac{C_{8,5}*C_{12,5}}{C_{20,5}} = 0.0036$

All 15 students are girls:

Groups are independent, so we multiply the probabilities:

$P = P_1*P_2*P_3 = 0.0163*0.00006*0.0036 = 3.52 \times 10^{-9}$

$3.52 \times 10^{-9} = 3.52 \times 10^{-7}\%$ probability that all the 15 students selected are girls