Jury Pick Problem Combinatorics Suppose that a jury of 12 persons is to be selected from a pool of 25 persons who were called for jury duty. The pool comprises 12 retired persons, 6 employed persons, 5 unemployed persons, and 2 students. Assuming that each person is equally likely to be selected, answer the following:


*

*What is the probability that both students will be selected?

*What is the probability that the jury will contain exactly twice as many retired persons as employed persons?


Solutions:


*

*$0.22$

*$0.1358$


Unfortunately I was not provided with an explanation of why these are the solutions to the problem and I couldn't find the proper textbook solutions online. What My reasoning in answering the first has been so far (wrong unfortunately):


*

*$\qquad\dfrac{\binom{25}{2}\cdot \binom{24}{1} \cdot \binom{23}{10}}{\binom{25}{12}}$


And I don't understand what exactly is wrong with this procedure. If anyone could also give me further advice on how to approach these problems it would be extremely useful!
Thank you!
 A: As regards the first question the answer is simply
$$\frac{\overbrace{\binom{2}{2}}^{\text{students}}\cdot\overbrace{\binom{23}{10}}^{\text{not students}}}{\binom{25}{12}}.$$
For the second question there are a few cases to consider. The jury of $12$ persons can be selected in one of the following ways:
$$(\text{retired},\text{employed},\text{others})\in\{(8,4,0),(6,3,3),(4,2,6),(2,1,9)\}.$$
Then the probability that one of these distinct cases happens is
$$\frac{\overbrace{\binom{12}{8}\binom{6}{4}\binom{7}{0}}^{(8,4,0)}+
\overbrace{?}^{(6,3,3)}
+\overbrace{?}^{(4,2,6)}+\overbrace{?}^{(2,1,9)}}{\binom{25}{12}}.$$
Can you take it from here?
A: 25C2 * 24C1 is presumably your attempt and finding the number of ways of choosing students, that doesn't really make sense. Once you've chosen two students, you're done, so you shouldn't have two expressions. Furthermore, you are choosing two students out of two students, so it should be 2C2, or just 1; there is only one way to choose two students.
Another way to calculate it: there are 12 choices as to where to put the first student, and 11 for the second, for a total of 132 ways to choose spots for the students. The probability of choosing a student for the first spot is 2/25, and for the second one, it's 1/24. This would give you (12*11)(2/25)(1/24) = .22 chance of getting both students.
