How do I calculate this probability? A student must answer $10$ questions, for each question $4$ propositions are given. If $X $is the number of correct questions, what is the probability of $5$ correct answers $\mathbb P(X=5)$?
Should we just divide $5/10$? 
if yes what about the $40$ propositions?
Should we devide $5/40$?
but we know that there is exactly $10$ correct answers.
What I'm missing here?
 A: Each question has $4$ choices so total outcomes are$4^{10}$
Now you can have five right answers in ${10}\choose 5$ ways 
So probability is $\frac{{10}\choose 5}{4^{10}}$
A: We use Binomial Distribution here. The formula that finds the probabilities for the binomial distribution for probability of success $p$, fixed number of trials $n$, and $k$ successes is as follows: $$\binom{n}{k}p^k(1-p)^{n-k}$$ Here $p$ stands for the probability of a correct answer and obviously $(1-p)$ for that of a wrong answer. Let $X$ denote the variable to represent the number of total correct answers in the quiz. We need to calculate $P(X=5)$. Using $p=0.25$ as there is equal probability of choosing any of the five options. Using the distribution formula, we get, $$P_{req} = \binom{10}{5}(0.25)^5(0.75)^5 = 0.058$$ Hope it helps.
A: I am using bionomial distribution to solve this question. 
$p( \text{correct answer}) = \frac{1}{4}$ 
As only 1 option correct out of 4.
Then,
$q( \text{incorrect answer}) = 1 - \frac{1}{4} = \frac{3}{4}$
Now according to bionamial distribution
$P(X=r) = C(n,r) (q)^{n-r} (p)^r$
Here X = 5 as correct answer and n = 10 total questions.
$P(X=5) = C(10,5) * \left(\frac{3}{4}\right)^5 * \left(\frac{1}{4}\right)^5 $
