The probability of leaving an umbrella in specific store A man with an umbrella visits three stores. In each store, the probability of leaving the umbrella is $\frac{1}{4}$. If the man did leave his umbrella, what is the probability that he left it in store 1, 2, and 3 respectively? This is my thoughts on the problem. Since the probaility of not leaving the umbrella in a given store is $\frac{3}{4}$, the probability of leaving it in store $j$ can be written as 1 - $(\frac{3}{4})^j$. My problem now is that I don't know how to calculate the conditional probabilities.

In general, the only way the man could have left his umbrella in the $k^\text{th}$ store is if he kept his umbrella in the first $(k-1)$ stores, then left it in the $k^{\text{th}}$ store. Hence, for each $k \in \\{1,2,3,4\\}$, it follows that: \begin{align*} &\text{Pr}[\text{man left it in the $k^\text{th}$ store} \mid \text{man left it in 1 of the 4 stores}] \\\ &= \frac{\text{Pr}[\text{man left it in the $k^\text{th}$ store AND man left it in 1 of the 4 stores}]}{\text{Pr}[\text{man left it in 1 of the 4 stores}]} \\\ &= \frac{\text{Pr}[\text{man left it in the $k^\text{th}$ store}]}{1 - \text{Pr}[\text{man never left it in any of the 4 stores}]} \\\ &= \frac{(\frac{3}{4})^{k-1}(\frac{1}{4})}{1 - (\frac{1}{4})^4} \\\ \end{align*}