Question

A class has fifteen boys and five girls. Suppose three students are selected at random from class. The probability that there are two boys and one girl is
A. \[\dfrac{{35}}{{76}}\]
B. \[\dfrac{{35}}{{38}}\]
C. \[\dfrac{7}{{76}}\]
D. \[\dfrac{{35}}{{72}}\]

Answer 1

Hint: First see what are the total no. of students in the class and how many are to be taken randomly out. For selecting we use a combination formula as \[{}^{n}{C_r}\]= \[\dfrac{n!}{{(n-r)!}\ {r!}} \] which represents selecting r things randomly out of n things. Since it's two boys and one girl so there can be many possible outcomes. Find out that and then the probability of having two boys and one girl.

Complete step-by-step solution -
Given that the class has 15 boys and 5 girls. So the total no. of students = 20 and we have to select out only 3 members. So, the combination becomes \[{}^{20}{C_3}\]= \[\dfrac{{20 \times 19 \times 18}}{{3 \times 2}}\] = 1140
Three students are selected at random. To find the probability that there are two boys and one girl can be done as follow
Selection of two boys out of 15 can be done as \[{}^{15}{C_2}\] = \[\dfrac{15!}{{13!}\ {2!}}\] = \[\dfrac{{15 \times 14}}{2}\] = 105
Similarly, selection of one girl out of 5 can be done as \[{}^5{C_2}\] = \[\dfrac{5!}{{3!}\ {2!}}\] = \[\dfrac{5 \times 4}{2}\] = 10

Hence, the required probability is \[\dfrac{{{}^{15}{C_2} \times {}^5{C_1}}}{{1140}}\] = \[\dfrac{{105 \times 10}}{{1140}}\] = \[\dfrac{{35}}{{38}}\]
∴ the correct option is ‘A’

Note: Key concept of such types of problems is to find out all the favourable combinations which are possible. Here we use the multiplication method of probability as the selection of 3 members are dependent on each other.