Original question: In your class at school there are 7 students of the 20 that wear glasses. If a group of 5 students is selected at random to be part of "Recycling committee", what is the probability that all will be wearing glasses?
Work:
P(All 5 in the committee are wearing glasses) = # of events when this is true/total # of events
Total # of events = $\frac{n!}{r!(n-r)!}$ = $\frac{20!}{5!(20-5)!}$
I cannot seem to deduce the # of events when this is true.
Is it simply $\frac{7}{20}$?
On
The total number of $5$-person committees you can choose would be $\binom{20}{5}=\frac{20!}{5!15!}$. The total number of "all-glasses" committees would be the same idea, but working only with the pool of glasses-wearers. Thus, the numerator of your probability fraction should be $\binom75=\frac{7!}{5!2!}$.
On
This is a special case of the hypergeometric distribution. There are $\binom{20}{5}$ groups of five people each having the same probability of being chosen. Of these groups, $\binom{7}{5}\binom{13}{0}$ groups have all $5$ people wearing glasses (the five people who wear glasses must be chosen among the seven in the class who wear glasses). Thus the probability is $$ \frac{\dbinom{7}{5}}{\dbinom{20}{5}} $$
HINT:
The number of ways to choose $5$ out of those $7$ students wear classes: $\dfrac{7!}{5!2!}=21$ ways.
The number of ways to choose $5$ out of all $20$ students: $\dfrac{20!}{5!15!}=15504$ ways, is this enough?