Probability of selecting different kinds of items from a set of items A debating committee of 5 students is to be selected at random by the instructor of a political science class consisting of 14 girls and 16 boys. Determine the probability of the following events:


*

*the committee will consist of all boys

*the committee will consist of $3$ girls $2$ boys

*the committee will consist of $5$ girls.


My answers:


*

*all $5$ boys means $\frac{16}{30} \times \frac{15}{29} \times \frac{14}{28} \times \frac{13}{27} \times \frac{12}{26}$

*the possibilities of having $3 $g and $2 $b are $9$ (gggbb, ggbbg, etc) so $\frac{9}{25}$

*is this the same as 1 but for girls? 

 A: Your answer to $(1)$ is correct, and you can use the same process for $(3)$: but the numerators change: 
For $(3)$, the probability of selecting an "all-girl" committee of five will be:
$$(3)\quad \frac{14}{30} \times \frac{13}{29} \times \frac{12}{28} \times \frac{11}{27} \times \frac{10}{26}$$
For $(2)$, you're off by one: there are $10$ patterns, and the probability of each varies.
You're better off approaching $(2)$ using binomial coefficients: it greatly simplifies the process, and is less prone to error than just brute force analysis.
To use binomial coefficients for $2$, note that there are $\text{(baseline)}\; = \displaystyle \binom {30}{5}$ equally likely ways of choosing a committee of five students from 30. And of those, there are $\displaystyle (\text{3 girls, 2 boys})\;=\;\binom{14}{3} \cdot \binom{16}{2}$ ways of choosing a committee consisting of exactly 3 girls and 2 boys.
Then, the probability of selecting such a committee is equal to $$\frac{\text{3 girls, 2 boys}}{\text{baseline}} = \frac{\binom{14}{3}\binom {16}{2}}{\binom{30}{5}}$$
