I've a problem with this exercise:
A program has to use $8$ processors from $A_1,...,A_{10}$.
$a)$ Find the probability that $A_1$ is used.
$b)$ The program has to use $8$ processors from $A_1,...,A_{30}$. Find the probability that exactly two from $A_1,A_2,A_3,A_4$ are used.
MY TRY:
$a)$ It should be like that: $$\binom{10}{8}-\binom{9}{8} = 36 \longrightarrow \frac{36}{45} = \color{red}{0.8}$$ $b)$ I need help to solve this question.
$a)$ is correct. If $8$ processors are used, the probability that $A_1$ is one of them is simply $\frac{8}{10}$
Hint for
$b)$ : We have $\binom{30}{8}$ possibilities to choose $8$ processors out of the $30$. We have $\binom{4}{2}\cdot \binom{26}{6}$ possibilities, if exactly two processors are under $A_1\cdots A_4$
The distribution producing these probabilities is called "hypergeometric distribution". It is useful to analyze Lotto or similar games.
a) Total combinations possible is $\binom{10}{8}$. And number of combinations possible given $A_1$ to be picked for sure is $\binom{9}{7}$. So the corresponding probability is $$\frac{\binom{9}{7}}{\binom{10}{8}}.$$
b) Total combinations possible is $\binom{30}{8}$. And number of combinations possible given exactly 2 from $A_1,A_2,A_3,A_4$ to be picked is $\binom{4}{2} \cdot \binom{26}{6}$. So the corresponding probability is $$\frac{\binom{4}{2} \cdot \binom{26}{6}}{\binom{30}{8}}.$$