Consider a class with 100 students enrolled. Suppose that 30 achieved a mark over 70%,60 achieved between 50-69% and 10 achieved 0−49% . Let's take a randomly selected sample of 12 of these students for course moderation purposes such that they can be considered independent of each other.
(a) Write down the probability distribution function of the number receiving each of the three type of grades $X=(x_1,x_2,x_3)$. Remember to note the constraints on the sample space.
(b) Compute the probability of the sample containing 6 with grade A, 4 with grade B+ and 2 with grade C
(c) Compute the probability of the sample containing 6 with grade A
Could someone help with part (c) of this question?
I was taking the following approach but am unsure if it's correct. It seems wrong to me and I feel as though I'm missing logic.
$P(X_1 = 6) = \frac{12!}{6!\cdot 0!\cdot 6!} \cdot 0.3^6 \cdot 0.6^0 \cdot 0.1^6 = .00005$
You have to calculate the sum of all probabilities where $x_1=6$ and $x_1+x_2+x_3=12$. This is
$$P(X_1=6)=\sum\limits_{x_2=0}^{6} \frac{12!}{6!\cdot x_2!\cdot (12-6-x_2)!}\cdot 0.3^6\cdot 0.6^{x_2}\cdot 0.1^{12-6-x_2}$$
For instance: If $x_2=2$ and the given $x_1=6$, then $x_3=12-6-2=4$. The sum is $12$. You add up all combinations, where $x_2+x_3=6$.
$$=\frac{12!/6!}{6!}\cdot 0.3^6\cdot 0.1^{6}\cdot \sum\limits_{x_2=0}^{6}\frac{6!}{x_2!\cdot (12-6-x_2)!}\cdot 6^{x_2}$$
$$=\frac{12!}{6!\cdot 6!}\cdot 0.3^6\cdot 0.1^{6}\cdot \sum\limits_{x_2=0}^{6} \binom{6}{x_2}\cdot 6^{x_2}\cdot 1^{6-x_2}$$
For the sum you can use the binomial theorem.
Update
More general: For arbitrary $p_i,x_1$ and $\sum\limits_{i=1}^{3} x_i=n$ the probability is
$$P(X_1=x_1)=\frac{n!}{x_1!\cdot (n-x_1)!}\cdot p_1^{x_1}\cdot (p_2+p_3)^{n-x_1},$$
where $p_2+p_3=1-p_1$.Therefore the marginal pmf is distributed as $X_1\sim \textrm{Bin}(n,p_1)$.