One way to approach this would be to break down $X$, the number of trials required, into the sum $X = X_1 + X_2 + X_3 + X_4$, where $X_i$ is the number of trials needed to get the $i^{th}$ unique coupon after having already drawn $i-1$ unique coupons.
It is relatively simple to evaluate $E[X_i]$ for each $i$:
- $X_1 = 1$ deterministically, since no matter what coupon you draw in the first envelope, it will be the first unique coupon type you have drawn. Therefore, $E[X_1] = 1$.
- Once you have drawn 1 unique coupon, there is a 75% probability that each new envelope you have will have a new type of coupon and a 25% probability that each new envelope you have will be the type of coupon you already have. Therefore, the number of envelopes you need to draw before you get your second unique coupon is distributed as $X_2\sim \mathrm{Geom}(0.75)$, where $\mathrm{Geom}(p)$ is the geometric distribution that counts the number of trials before the first success, where the probability of success is $p$. Therefore $E[X_2] = 1/0.75 = 4/3$.
- Similarly to the logic above, we see $X_3\sim \mathrm{Geom}(0.5)$ and $X_4\sim \mathrm{Geom}(0.25)$, meaning $E[X_3] = 2$ and $E[X_4] = 4$.
Pulling this all together, we have $E[X] = E[X_1 + X_2 + X_3 + X_4] = 25/3$. Therefore the expected expenditure is Rs. 250.
self-studytag and read its wiki. You should also edit your question to include what you have done and where you are stuck $\endgroup$ – Marquis de Carabas Apr 25 '16 at 1:41