Question

From a group of 10 men and 5 women, four member committees are to be formed each of which must contain at least one woman. Then the probability for these committees to have more women than men is\[\]
A.$\dfrac{3}{11}$\[\]
B. $\dfrac{2}{23}$\[\]
C. $\dfrac{1}{11}$\[\]
D. $\dfrac{21}{220}$\[\]

Answer 1

Hint: Use the formula for combination and fundamental principle of counting to find out the combinatorial values for both the cases.the find the probability.

Complete step by step answer:
We know that the selection of $r$ entities from $n$ unique entities is given by $^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$. We also know by fundamental principle of counting, simultaneous selection of $r$ entities from $n$ and $q$ entities from $m$ is given by $^{m}{{C}_{q}}^{n}{{C}_{r}}$. \[\]
Here we infer from the question that the number of women is given by $n=5$ and the number of men is given by $m=10$. A committee of 4 has to be formed. So the possible combinations are\[\]
(i) Out of 4 members there is one $r=1$ is woman and the rest 3$(q=3)$ are men So total such groups are $^{5}{{C}_{1}}^{10}{{C}_{3}}$ \[\]
(ii) Out of 4 members there are 2 women and the rest 2 are men. So total such committees are $^{5}{{C}_{2}}^{10}{{C}_{2}}$\[\]
iii) Out of 4 members there are 3 women and the rest one is man . So total such committees are $^{5}{{C}_{3}}^{10}{{C}_{1}}$\[\]
iv) Out of 4 members all are women. So total such committees are $^{5}{{C}_{4}}^{10}{{C}_{0}}$\[\]
So total number ways four member committee with at least one woman can be formed =\[^{5}{{C}_{1}}{{\times }^{10}}{{C}_{3}}{{+}^{5}}{{C}_{2}}{{\times }^{10}}{{C}_{2}}{{+}^{5}}{{C}_{3}}{{\times }^{10}}{{C}_{1}}{{+}^{5}}{{C}_{4}}^{10}{{C}_{0}}=600+450+100+5=1155=S\left( \text{say} \right)\]
The possible ways committees to have more women than men is\[\]
(i) There are 3 women and 1 man. . So total such committees $^{5}{{C}_{2}}^{10}{{C}_{1}}$\[\]
(ii) There are only women. . So total such committees $^{5}{{C}_{4}}$\[\]
So total number ways four member committee with to have more women than men can be formed is $^{5}{{C}_{2}}^{10}{{C}_{1}}{{+}^{5}}{{C}_{4}}=100+5=105=A(\text{say})$\[\]
So the probability of the committee being made with more women than men is $\dfrac{n\left( A \right)}{n\left( S \right)}=\dfrac{105}{1155}=\dfrac{1}{11}$\[\]

So, the correct answer is “Option C”.

Note: In this problem which tests the concept of combination, the proper intuitive use of selection formula and careful summation helps to arrive at the correct result. The question can be framed as the probability of a group being made with more men than women.