There are 50 bars in Beer City. Every weekend the Ancient Society of Beer Lovers randomly chooses 3 bars to visit. Assuming that the choices every weekend are independent, calculate expected value and variance in number of bars visited at least once during 10 consecutive weekends.
I have literally no idea how to tackle this and it seems pretty basic example problem of variance/expected value. Could anyone help?
The number of bars visited at least once is $50$ minus the number of bars never visited, so we can calculate the expected value and variance of the latter.
Let $I_k$ be an indicator variable that is $1$ if bar $k$ is never visited and $0$ otherwise. Then we want to know the expected value and variance of $S=\sum_kI_k$.
The probability for one particular bar to never be visited is $p_1=\left(\frac{\binom{49}3}{\binom{50}3}\right)^{10}=\left(\frac{47}{50}\right)^{10}$. The probability for two particular bars to never be visited is $p_2=\left(\frac{\binom{48}3}{\binom{50}3}\right)^{10}=\left(\frac{47\cdot46}{50\cdot49}\right)^{10}=\left(\frac{1081}{1225}\right)^{10}$. Thus, the expected value is
\begin{eqnarray} E[S] &=& E\left[\sum_kI_k\right] \\ &=& \sum_kE\left[I_k\right] \\ &=& \sum_kp_1 \\ &=& 50p_1 \\ &=& 50\left(\frac{47}{50}\right)^{10} \\ &=& \frac{47^{10}}{50^9} \\ &=& \frac{52599132235830049}{1953125000000000} \\ &\approx& 26.93 \end{eqnarray}
and the variance is
\begin{eqnarray} \operatorname{Var}[S] &=& E\left[\left(\sum_kI_k\right)^2\right]-E\left[\sum_kI_k\right]^2 \\ &=& \sum_{j,k}E[I_jI_k]-\sum_{j,k}E[I_j]E[I_k]) \\ &=& \sum_{j=k}E[I_jI_k]+\sum_{j\ne k}E[I_jI_k]-\sum_{j,k}E[I_j]E[I_k]) \\ &=& 50p_1+50(50-1)p_2-(50p_1)^2 \\ &=& 50(50-1)p_2-50p_1(50p_1-1) \\ &=& 50(50-1)\left(\frac{1081}{1225}\right)^{10}-\frac{47^{10}}{50^9}\left(\frac{47^{10}}{50^9}-1\right) \\ &=& \frac{20007424621745238891781620924108115919925351951}{6211904899255558013916015625000000000000000000} \\ &\approx& 3.22\;. \end{eqnarray}