Exponential r.v.s. problem

Question

Here is the question.

A total of n people has been invited to a party for honouring a scholar. The party begins at time 0. The arrival times of n guests are independent exponential r.v.s. with mean 1, and the arrival time of scholar is uniform distributed between 0 and 1. Find the probability that exactly k of the guests arrive before the scholar.

For my understanding, Let X be people arrive party, $0 \le x \le n$ and Y be the arrival times of x guest

$pY \mid X=n\{Y=t\mid X=n\} = λe^{−λx}$ for $x\gt 0$, then given mean, $E[Y] = \frac{1}{\lambda} = 1$ i.e. $\lambda=1$.

So $pY \mid X=n\{Y=t\mid X=n\}$ becomes $e^{−x}$ as we have to consider the time before scholar, so $0 \lt t \lt 1$. and $X=k$.

I know that I have to find $pX(X=k)$, which can be used the equation of $\frac{pYX}{pY}\mid X$. but how can I find $pYX$? or my concept is something wrong here?

Please help!!

Answer 1

A possible approach:

• Suppose the scholar arrives at time $t$:
  • What is the probability a particular guest arrives before time $t$?
  • What is the probability a particular guest arrives after time $t$?
  • What is the probability exactly $k$ out of $n$ guests arrive before time $t$?
• What happens when you combine this with the scholar's arrival time being uniformly distributed?
  • Hint: $\displaystyle P(K=k) = E_T[P(K=k\mid T=t)]=\int_t P(K=k \mid T=t)\, p(t)\, dt$