Largest value of $P(X =4)$ Consider $4$ trials each having same probability of success. Let $X$ denote the total number of successes in these trials, if $\mathbb{E}(X) = 2$, what is 
A) the maximum value of $P(X =4)$
B) the minimum value of $P(X = 4)$
I don’t really know how to approach this question, all i did was say that
$\mathbb{E}(X) = np = 4p =2$
Which gives $p =\frac{1}{2}$ so $P(X =4)=\frac{1}{16}$ but this is not correct and I don’t know how to do the second part either
 A: Even though my answer does not agree with the one given in your book, I will post it here, to see if someone else in the community finds the error in my approach

If you call $S_i$, $1\leq i \leq 4$ the event of success in the $i$-th trial, then
$$P(S_1 \cap S_2 \cap S_3 \cap S_4) = P(S_1) \cdot P(S_2|S_1) \cdot P(S_3|S_1\cap S_2) \cdot P(S_4|S_1\cap S_2 \cap S_3).$$
You already noted that $P(S_i) = \frac{1}{2}$, so we must have
$$0 \leq P(S_1 \cap S_2 \cap S_3 \cap S_4) \leq \frac{1}{2}.$$

Let us see two scenarios in which extreme values are attained.


*

*Two urns containing $4$ black balls and $4$ white balls, respectively.

*Two urns, one containing $1$ black ball and $3$ white balls, one containing $3$ back balls and $1$ white ball.


In both cases, pick randomly one urn and extract $4$ balls and let $X$ be the random variable counting the number of, say, black balls extracted. 
In the first scenario $P(X = 0) = P(X=4) = \frac{1}{2}$ and $P(X=1) = P(X=2) = P(X=3) = 0$.
In the second scenario $P(X=0) = P(X=2) = P(X=4) = 0$ and $P(X=1) = P(X=3) = \frac{1}{2}$.
So in both cases $\Bbb E[X]=2$.
Let us now check $P(S_i)$, which is given by
$$P(S_i) = \frac{P(S_i|U_1) + P(S_i|U_2)}{2},$$
$U_1$ and $U_2$ being the event of picking first and second urn respectively.
In the first scenario $P(S_i|U_1) = 1$ and $P(S_i|U_2) = 0$, so that $P(S_i) = \frac{1}{2}.$
In the second scenario $P(S_i|U_1) = \frac{1}{4}$ and $P(S_i|U_2) = \frac{3}{4}$. Thus, again $P(S_i) = \frac{1}{2}$ as required.
