Question 322522
<font face="Garamond" size="+2">

The probability of *[tex \Large k] successes in *[tex \Large n] trials where *[tex \Large p] is the probability of success on any given trial is given by:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ P_n(k)\ =\ \left(n\cr k\right\)p^k\left(1\,-\,p\right)^{n\,-\,k}]


Where *[tex \LARGE \left(n\cr k\right\)] is the number of combinations of *[tex \Large n] things taken *[tex \Large k] at a time and is calculated by *[tex \Large \frac{n!}{k!(n\,-\,k)!}]


Here you need 3 successes in 3 trials where the probability of success is 0.4 (4 out of 10 names were women)


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ P_3(3)\ =\ \left(3\cr 3\right\)(0.4)^3(0.6)^{0}]


The arithmetic is yours to do.  Hint:


*[tex \LARGE \left(n\cr n\right\)\ =\ 1\ \forall\ n\ \in\ \mathbb{Z},\ n\ >\ 0] 


and


*[tex \LARGE x^0\ =\ 1\ \forall\ x\ \in\ \mathbb{R}]



John
*[tex \LARGE e^{i\pi} + 1 = 0]
</font>