Finding probability by using combinations There are 20 doctors and 15 engineers attending a conference. The number of women doctors and that of women engineers are 12 and 5 respectively. Four participants from this group are selected randomly to chat some sessional panel discussion. Given that two women are selected, find the probability that both of them are doctors. So I approached the question with $$P(\text{Both of the women are doctors}|\text{Both are women}) = \frac{P(\text{Both women doctors AND Both are women})}{P(\text{Both are women})}$$ $$P(\text{Both women doctors AND both are women}) = \frac{\binom{12}{2}\times\binom{23}{2}}{\binom{35}{4}},$$ $$P(\text{Both are women}) = \frac{\binom{17}{2} \times \binom{18}{2}}{\binom{35}{4}}$$ and afterwards, I'll substitute back the answer in the first equation. Is my method of approaching this question, correct?

$4$ persons are selected out of $12$ female doctors, $5$ female engineers and $18$ men.

Under the original condition that exactly $2$ women are selected there are$\binom{17}{2}\binom{18}2$ possibilities.

Under the extra condition that the $2$ women are both doctors there are $\binom{12}2\binom{18}2$ possibilities.

So the probability that the extra condition will be satisfied under the original condition is: $$\frac{\binom{12}2\binom{18}2}{\binom{17}{2}\binom{18}2}=\frac{\binom{12}{2}}{\binom{17}{2}}=\frac{12\cdot11}{17\cdot16}=\frac{33}{68}$$