$$ \text { A group has } 20 \text { Actuaries and } 11 \text { Biostaticians in it. All } 31 \text { people look different (no twins!). } $$
a. $$ \begin{array}{l}{\text { I asked everyone in the group to send me an email. As it happened, ALL the Actuaries }} \\ {\text { responded before ANY of the Biostaticians did. What was the probability of such a result, if }} \\ {\text { all these people have similar email habits? }}\end{array} $$
b. $$ \begin{array}{l}{\text { If we choose } 7 \text { persons at random from the group, find the probability that exactly } 4 \text { of them }} \\ {\text { will be Actuaries. }}\end{array} $$,
Part b is already done by me but for part a I am unable to understand it properly can somebody help me regarding that.
If all people have the same habit of replying, then each sequence of $31$ people is equally likely to occur. In our case, however, it turns out that all actuaries reply before all biostaticians. There are $20!$ ways to sort the former, while there are $11!$ ways to sort the latter. The total number of valid sequences thus equals $20! 11!$, and since each sequence is equally likely, the probability of such result equals:
$$\frac{20!11!}{31!} \approx 1.18 \cdot 10^{-8}$$
Alternatively, look at it this way. The probability of the first reply being from an actuary, equals $\frac{20}{31}$. Then, the probability of the second reply being from another actuary, equals $\frac{19}{30}$. Extending this to all actuaries, we find a probability of:
$$\frac{20}{31} \frac{19}{30} \frac{18}{29} \ldots \frac{2}{13} \frac{1}{12} \frac{11}{11} \frac{10}{10} \ldots \frac{1}{1} = \frac{20!11!}{31!}$$