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
On
If you treat each grade as an independent event, the probability of no students earning an A is $(0.7)^{15}$. The probability of only one student earning an A is $15(0.3)(0.7)^{14}$
The probability of at least two students having A's is $1 - \Pr (0 A's) - \Pr( 1 A)$.
$\Pr(2+ A) = 1 - (0.7)^{15} - 15(0.3)(0.7)^{14}$
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 $$1-\left((0.7)^{15}+15(0.3)(0.7)^{14}\right).$$