Suppose that $X$ is a random variable with Bernoulli distribution $B_p$ with probability parameter $p$.

Assume that the variance $V(X) = 0.21$. We further assume that $p > 0.5$.

(a) Find the probability $p$.

(b) Find the expectation $E(X)$.

Hint.

Recall that if $X$ is a Bernoulli random variable with parameter $p$, then the expectation $E[X]$ and the variance $V(X)$ of $X$ are given by
\begin{align*}
E[X] &= p \\
V(X) &= p(1-p).
\end{align*}

For proofs of the formulas, see that post Expectation, Variance, and Standard Deviation of Bernoulli Random Variables.

Solution.

Solution of (a)

Recall (see the hint above) that the variance of a Bernoulli random variable $X$ with parameter $p$ is
\[V(X) = p(1-p).\] Thus we have
\[p(1-p) = 0.21.\] This yields the quadratic equation
\[p^2-p+0.21 = (p-0.3)(p-0.7) = 0.\] Solving the equation, we obtain $p = 0.3$ or $p = 0.7$. By assumption, $p > 0.5$. Hence we conclude that $p = 0.7$.

Solution of (b)

The expectation $E(X)$ of a Bernoulli distributed $X$ is given by $E(X) = p$ (see the hint above). As we obtained $p=0.7$ in Part (a), we see that the expectation is $E(X) = 0.7$.

The post Given the Variance of a Bernoulli Random Variable, Find Its Expectation first appeared on Problems in Mathematics.