Joint exponential probability 
This is from gate 2021 statistics paper. The answer in key is given as 143. I got answer as 4 .
I used the formula for joint distribution under the condition $x_1+x_2 < y$
Here is my work:
if we let x1 and x2 as the arrival times for the next two visitors  while y is the departure time for one of the 10 visitors.
$x_1 \sim  exp(1)$
$x_2 \sim exp(1)$
$y \sim  exp(1)$
$x_1+x_2 = X \sim  erlang(1)$
X < y,  X >0
If we perform a double integral with above conditions , it will give 1/4, so 1/p = 4
$p =\int _{ X=0 }^{ \infty }\int _{ y= X }^{ \infty }xe^{-X}.e^{-y}\, dx dy = 1/4$
 A: Let $X$ be the time of the next departure of the 10 people, then $X$ is the minimum of 10 Exponential(1) random variables and so is distributed: $X\sim \text{Exponential}(10)$.
Let $Y$ be the number of arrivals by the next departure, then $Y$ given the next departure time $X$ is distributed as $Y|X\sim \text{Poisson}(X)$.
We are interested in $\Pr(Y=2)$ which by the law of total probability can be found to be
$$\begin{split}\Pr(Y=2)&=\int_0^\infty \Pr(Y=2|x)f(x)dx\\
&=\int_0^\infty \frac{e^{-x} x^2}{2!}10e^{-10x}dx\\
&=\int_0^\infty 5e^{-11x}x^2dx\end{split}$$
After doing integration by parts twice we get that $p=\Pr(Y=2)=\frac{10}{1331}$ so $\frac 1p=133.1$.
Here is an alternative solution that gives the same answer. Let $X$ be the arrival time of the next two people, $Y$ be the departure time of the next person out of the 10 (thanks to the memorylessness property), and $Z$ be the arrival time of the third person.
$$\begin{split}X&\sim\text{Gamma}(2,1)\\
Y&\sim\text{Exponential}(10)\\
Z&\sim \text{Exponential}(1)\end{split}$$
We are interested in $\Pr(Z>Y-X,X<Y)$, because we need two people to arrive before the 1st departure and the third person to arrive after it. This integral is
$$\int_0^\infty\int_x^\infty\int_{y-x}^\infty10xe^{-(x+10y+z)}dzdydx\\
=\int_0^\infty\int_x^\infty 10xe^{-11y}dydx\\
=\int_0^\infty \frac{10xe^{-11x}}{11}dx\\
=\frac {10}{1331}$$
Conclusion: Answer key is wrong. This problem is also intractable without consulting the memorylessness property.
A: I'm a bit late to the party, but I thought someone may find it useful to know that the accepted answer isn't correct. Indulge me while I demonstrate my approach.
First suppose there are $n\geq 1$ in attendance at the amusement park. The probability that someone arrives before anyone in the park leaves is $$\mathbb{P}\left(\min\{X_1,...,X_n\}>Y\right)=\frac{1}{n+1}$$ Here $X_1,...,X_n\sim \text{Exponential}(1)$ represent the times each of the n individuals spend at the amusement park and $Y\sim \text{Exponential}(1)$ is the time is takes for the next individual to arrive at the amusement park.
Since we're starting with $10$ individuals in the amusement park and the exponential distribution is memoryless, the probability that exactly two individuals leave before the next customer arrives is $$p=\frac{1}{11}\cdot \frac{1}{12}\cdot \left(1-\frac{1}{13}\right)=\frac{1}{143}$$ Hence $\frac{1}{p}=143$
