Question

Mr. Adams has $14$ boys and $8$ girls in his mathematics class. If he chooses two students at random to work on the board, what is the probability that both students chosen are boys?

Answer 1

Step-1: Write the information given in the question:

The total number of boys in a class is $14$.

The total number of girls in a class is $8$.

Since the total number of students in the class is the sum of the boys and girls in the class

Therefore, the total number of students in the class is $14+8$, i.e. $22.$

Step-2: Find the probability that both the chosen students are boys:

The probability of an event is obtained by dividing the number of favorable outcomes by the total number of outcomes.

If the two kids are chosen and both are boys, then the number of a favorable outcome is ${}^{14}C_2.$

The total number of outcomes is ${}^{22}C_2.$

Therefore, the probability of choosing both the boys is

$\begin{array}{rcl}P& =& \frac{{}^{14}C_2}{{}^{22}C_2}\\ & =& \frac{{\displaystyle \frac{14!}{2!12!}}}{{\displaystyle \frac{22!}{2!20!}}}\\ & =& \frac{{\displaystyle \frac{14\times 13\times 12!}{2!12!}}}{{\displaystyle \frac{22\times 21\times 20!}{2!20!}}}\\ & =& \frac{{\displaystyle \frac{14\times 13\times 12!}{2!12!}}}{{\displaystyle \frac{22\times 21\times 20!}{2!20!}}}\\ & =& \frac{{\displaystyle \frac{14\times 13}{2!}}}{{\displaystyle \frac{22\times 21}{2!}}}\\ & =& \frac{14\times 13}{2!}\times \frac{2!}{22\times 21}\\ \therefore & =& \frac{13}{33}\end{array}$

Hence the probability that both chosen students are boys is $\frac{13}{33}.$