Solving for $2$ successes in a binomial distribution wit $E(X)$ and $E(X^2)$ known.

Question

A random variable $X$ follows a binomial probability distribution with probability $P(X)$, with mean as $2$, probability of success as $p$ and probability of failure as $q$ such that $p + q = 1$. If $Σx^2P(x)=28/5$, then the probability of exactly $2$ success is what?

Answer 1

This question requires you to understand binomial probability. In this situation we have $2$ unknowns, $n$, (num of trials), and $p$, (the probability of success). As such, we need $2$ equations. The first one is given, such that the mean is $2$, in a binomial distribution, the mean, $\mu$, is given by:

$$\mu = np$$ $$np = 2$$

For the second equation, we can calculate the variance of the distribution, which is given by:

$$Var(X) = E(X^2)-E(X)^2$$ $$Var(X) = \frac{28}{5}-4 = \frac{8}{5}$$

Variance in a binomial distribution is also given by $np(1-p)$. So, for our second equation we can derive:

$$np(1-p)=\frac{8}{5}$$

Now, subbing in $np=2$:

$$2(1-p)=\frac{8}{5}$$ $$p = 0.2, n=10$$

The question asks for $2$ successes, so we use the binomial probability formula:

$$Pr(X=x) = \binom{n}{x} \times p^x \times (1-p)^{n-x}$$ $$Pr(X=2) = \binom{10}{2} \times 0.2^2 \times (1-0.2)^{10-2}$$

Which is approximately equal to $0.301989888$.