More tricky probability Problems 3 officials, the President, Secretary general and treasurer of the students body are to be elected from a group of 5 male and 7 female students. Find the probability that:
a) all officials are male
b) 2 of the officials are female
 A: The problem is unsolvable without further information.  Perhaps the male candidates are hopelessly bad, and no one will vote for them. 
We substitute a problem we can solve, probably the problem you atr intended to solve. A president, a secretary, and a treasurer, all distinct, are chosen at random from a group of $5$ males and $7$ females, with all choices equally likely. What is the probability that all the chosen people are male? What is the probability that exactly $2$ of the chosen people are female?
One way to solve the first problem is as follows. Imagine the selection is done in the order president, secretary, treasurer.
The probability the president is male is $\frac{5}{12}$. Given the president is male, the probability the secretary is male is $\frac{4}{11}$. And given these things have happened, the probability the treasurer is male is $\frac{3}{10}$. Thus the required probability is 
$$\frac{5}{12}\cdot\frac{4}{11}\cdot \frac{3}{10}.$$
Another way: There are $\binom{12}{3}$ ways to select the set of members of the executive (note we are not assigning positions). There are $\binom{5}{3}$ ways to select $3$ males. So our probability is 
$$\frac{\binom{5}{3}}{\binom{12}{3}}.$$
Now we solve the second problem. The easiest approach is by noting that there are $\binom{5}{1}\binom{7}{2}$ ways to choose $1$ male and $2$ females. So the required probability is
$$\frac{\binom{5}{1}\binom{7}{2}}{\binom{12}{3}}.$$ 
Another way: If we want $2$ female and $1$ male, we can have female president, secretary, and male treasurer, or female president, male secretary, female treasurer, or male president and female secretary and treasurer. We find the probability of each of these, and add.
For example, the probability of female president, female secretary, and male treasurer is $\binom{7}{12}\cdot \binom{6}{11}\cdot \binom{5}{10}$. We obtain two similar expressions for the other two patterns. (The numbers turn out to be the same.) 
