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.