At the Hairdresser

Let $X$ be the duration of the first appointment and $Y$ the duration of the second appointment.

We note that even though the second appointment is scheduled for 12:15, it cannot start until the first one ends. On the other hand, if the first one ends earlier than 12:15, it cannot start until 12:15. Therefore the duration $T$ from the noon until the end of the second appointment is given by $T = Z+Y$ , where $Z= \max\{15, X\}$

$\begin{aligned} \mathbb{E}[T] &= \mathbb{E}[Z] +\mathbb{E}[Y]\\ & = \mathbb{E}[Z] +30 \end{aligned}$

But the memoryless property of the exponential distribution implies that the expected additional time for the first appointment conditioned on being at least 15 minutes is still 30 minutes, so that $\mathbb{E}[Z| X\ge15] = 45$ .

$\begin{aligned} \mathbb{E}[Z] & = 15 \mathrm{Prob}(X<15) + \mathbb{E}[Z| X\ge15] \mathrm{Prob}(X\ge15)\\ & = 15 \left(1- e^{-\frac{15}{30}}\right) + 45 e^{-\frac{15}{30}}\\ & = 15 + 30 e^{-1/2}\\ & \approx 33.19592 \end{aligned}$

Thus $\mathbb{E}[T] \approx 63.19592$ and multiplying by $10^{4}$ and rounding gives $\fbox{631959}$