# Grade Probability based on Binomial Distribution

The student knows answers to 75 out of 100 questions in the exam. He picks 3 random questions of the 100 and answers. Three correct answers yield grade A, two correct answers yield grade B, 1 correct answer yields C and 0 yields D.

What's the probability that 2 out of 5 students will get A, 2 will get B and 1 will get C?

I am using the following approach:

since the student knows 75 out of a hundred questions. probability of getting a known question in the exam would be p = $$\frac{3}{4}$$

Using binomial distribution I have calculated the probability of getting grade A, B and C as follows:

P(A) = $${5 \choose 2} \times p^2 \times q^3 = \frac{27}{64} \approx 0.42$$

P(B) = $${5 \choose 2} \times p^2 \times q^3 = \frac{27}{64} \approx 0.42$$

P(C) = $${5 \choose 1} \times p^2 \times q^3 = \frac{9}{64} \approx 0.14$$

Now to get the probability of 2 grade A, 2 grade B and 2 grade C students out of 5 should be:

P(X) = $${5 \choose 2} p_A^2 (1 - p_A^3) \times {5 \choose 2} p_B^2 (1 - p_B^3) \times {5 \choose 1} p_C (1 - p_C^3) \approx 0.04$$

But this is reported as wrong. Can anyone suggest what could be the problem?

P(X) = $$\frac{n!}{n_1! \times n_2! \times n_3!} p_1^{n_1} \times p_2^{n_2} \times p_3^{n_3}$$
Now $$p_1 = 0.42; n_1 = 2 \\ p_2 = 0.42; n_2 = 2 \\ p_3 = 0.14; n_3 = 1$$
P(X) = $$\frac{5!}{2! \times 2! \times 1!} \times 0.42^2 \times 0.42^2 \times 0.14 \approx 0.13$$