How do we calculate the probabilty in the following situation? There are 300 in a class, 6 have perfect score on first midterm and 9 have perfect score on second midterm. One person will be chosen randomly from the class. If possible with the information given, find the chance the person has perfect score on both midterm. 
I think the chance cannot be calculated because we don't know the people who get perfect on both midterms. But I am not sure if I understood the question.
 A: It cannot be calculated, But consider the following two cases: 


*

*None of the $6$ people who have perfect score on the first midterm are among the $9$ people with perfect score on the second midterm. The desired probability $p$ is zero in this case.

*All $6$ people with perfect score have also perfect score in the second midterm. In this case, the desired probability is $$p=p_{\text{max}}=\frac{\binom{6}{1}}{\binom{300}{1}}$$
So the desired probability can be bounded as follows:$$0\le p\le p_{\text{max}}$$
A: Let $X$ denote the number of students with a perfect score on both terms, then:
$$\forall{n\in[0,6]}:P(X=n)=\frac{\binom{300}{n}\cdot\binom{300-n}{6-n}\cdot\binom{300-6}{300-9-n}}{\binom{300}{6}\cdot\binom{300}{9}}$$
And the probability of choosing a student with a perfect score on both terms is:
$$\sum\limits_{n=0}^{6}P(X=n)\cdot\frac{n}{300}=\sum\limits_{n=0}^{6}\frac{\binom{300}{n}\cdot\binom{300-n}{6-n}\cdot\binom{300-6}{300-9-n}}{\binom{300}{6}\cdot\binom{300}{9}}\cdot\frac{n}{300}=1.94\%$$
