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:

The group contains $$10$$ men and $$5$$ women, so in total $$15$$ people. Four-member committees are to be formed with the condition that each committee must contain at least one woman. We wish to find the probability that such a committee has more women than men.

When a committee of four has more women than men, the number of women $$W$$ is greater than the number of men $$M$$. Because $$W+M=4$$, the only possible distributions are:

$$\begin{aligned} W=3,\;M=1 \quad\text{or}\quad W=4,\;M=0 \end{aligned}$$

First, the size of the sample space is the total number of four-member committees that contain at least one woman. Start with the number of unrestricted four-member committees chosen from all $$15$$ people:

$$\binom{15}{4}=\frac{15\times14\times13\times12}{4\times3\times2\times1}=1365\qquad -(1)$$

The only committees that violate the “at least one woman” condition are the all-male committees. The number of such all-male committees is:

$$\binom{10}{4}=\frac{10\times9\times8\times7}{4\times3\times2\times1}=210\qquad -(2)$$

Subtracting equation $$-(2)$$ from equation $$-(1)$$ gives the size of the sample space:

$$1365-210=1155\qquad -(3)$$

Next, count the favourable committees.

Case 1: 3 women and 1 man.
Choose $$3$$ women out of $$5$$ and $$1$$ man out of $$10$$:

$$\binom{5}{3}\times\binom{10}{1}=10\times10=100\qquad -(4)$$

Case 2: 4 women and 0 men.
Choose $$4$$ women out of $$5$$:

$$\binom{5}{4}=5\qquad -(5)$$

Adding equations $$-(4)$$ and $$-(5)$$ gives the total number of favourable committees:

$$100+5=105\qquad -(6)$$

The required probability is the quotient of equations $$-(6)$$ and $$-(3)$$:

$$\frac{105}{1155}\qquad -(7)$$

Simplify the fraction by dividing numerator and denominator by $$105$$:

$$\frac{105\div105}{1155\div105}=\frac{1}{11}\qquad -(8)$$

Therefore, the probability that a four-member committee satisfying the “at least one woman” condition has more women than men is $$\dfrac{1}{11}$$.

Hence, the correct answer is Option C.