Question

A committee of 5 principles is to be selected from a group of 6 male principals and 8 female principals. If the selection is made randomly, find the probability that there are 3 female principals and 2 male principals.

Answer 1

Hint : Solve the question with the help of combinations for selecting $3$ female principals out of $8$ female principals and also for selecting $2$ male principals out of $6$ male principals. At last divide the product of both the probabilities with the total combinations of selecting $5$ principals out of $14$ principles.

Complete step-by-step answer :
Number of ways in which 3 female principals are selected from 8 female principals
$\Rightarrow {\;^8}{C_3}$= $\dfrac{{8!}}{{3!\left( {8 - 3} \right)!}}$
= $\dfrac{{8!}}{{3!\left( 5 \right)!}}$
= $\dfrac{{8.7.6.5!}}{{3!\left( 5 \right)!}}$ ( cancel the same terms from numerator and denominator)
= $\dfrac{{8.7.6}}{{3.2}}$ = 8.7 =56
$\Rightarrow$ Number of ways in which 2 male principals are selected from 6 male principals =
$^6{C_2}$= $\dfrac{{6!}}{{2!\left( {6 - 2} \right)!}}$
= $\dfrac{{6!}}{{2!\left( 4 \right)!}}$
= $\dfrac{{6.5.4!}}{{2!\left( 4 \right)!}}$ ( cancel the same terms from numerator and denominator)
= $\dfrac{{6.5}}{2}$ =15
Number of ways 5 people selected among 14 principles randomly
$^{14}{C_5}$ = $\dfrac{{14!}}{{5!\left( {14 - 5} \right)!}}$ $\dfrac{{840}}{{2002}}$
= $\dfrac{{14!}}{{5!\left( 9 \right)!}}$
= $\dfrac{{14.13.12.11.10}}{{5!}}$ (cancel the same terms from numerator and denominator)
= $\dfrac{{14.13.12.11.10}}{{5.4.3.2.1}}$
=2002
$\Rightarrow$ Probability of 3 female principals and 2 male principals = $\dfrac{{^6{C_2}{.^8}{C_3}}}{{^{14}{C_5}}}$
= $\dfrac{{15\times 56}}{{2002}}$
= $\dfrac{{840}}{{2002}}$

Note : Firstly for this question revise all the concepts regarding the factorials, permutations and combinations which will help to find the probability. Revise the definitions of probability and conditional probability also.