Question

There are 20 students in a class. 11 of them are Chinese, 4 of them are Malay, and 5 of them
are Indian. 5 students are to be chosen to represent the class in a contest. What is the
probability of a selection consisting of only two Chinese?

Answer 1

Total number of ways to choose 5 students out of 20 will be equal to $C(20,5)$, where C represents the Combination.

$C(20,5)=\frac{20!}{5!15!}=15504$

Now, the total number of combinations in which out of these 5 only 2 are Chinese will be as follows:

• Two Chinese and the remaining are from Indian and Malay students.

So, total number of ways to choose 5 students with only 2 Chinese will be equal to

$C(11,2) \cdot C(9,3)=\frac{11!}{9!2!} \cdot \frac{9!}{3!6!}=$

$=55 \cdot 84=4620$

Thus, the probability of a selection consisting of only two Chinese will be

$\frac{C(11,2) \cdot C(9,3)}{C(20,5)}=\frac{4620}{15504}=0.298$