How do i apply multinomial laws in this question? the Question is  i assume i have 15 students in class 
A grade obtain probaiblity = 0.3   
B grade obtain probability =0.4
C grade obtain probability = 0.3
and I have this question 
What is the probabilty if we are given   2 students at least obtain A ?
Do I need apply this law ?
P=n!/((A!)(B!)(C!)) 〖P1〗^A 〖P2〗^(B ) 〖P3〗^C


We reword the question, perhaps incorrectly. 
If a student is chosen at random, the probabilities she obtains an A, B, and C are, respectively, $0.3$, $0.4$, and $0.3$.
If $15$ students are chosen at random, what is the probability that at least $2$ of the students obtain an A?
The probability that $a$ students get an A, $b$ get a B, and $c$ get a C, where $a+b+c=15$, is indeed given by the formula quoted in the post. It is
$$\frac{15!}{a!b!c!}(0.3)^a(0.4)^b (0.3)^c.$$
However, the above formula is not the best tool for solving the problem. 
The probability a student gets an A is $0.3$, and the probability she doesn't is $0.7$. We are only interested in the number of A's, so we are in a binomial situation.
We want the probability there are $2$ or more A's. It is easier to first find the probability of fewer than $2$ A's. This is the probability of $0$ A's plus the probability of $1$ A. 
If $X$ is the number of A's,
$$\Pr(X\lt 2)=\binom{15}{0}(0.3)^0(0.7)^{15}+\binom{15}{1}(0.3)^1(0.7)^{14}.$$
It follows that the probability of $2$ or more A's is
$$1-\left((0.7)^{15}+15(0.3)(0.7)^{14}\right).$$ 
