Probability - exponential distribution question I got the following question:
There are two cashiers in a supermarket, the waiting time for each is of exponential distribution with parameters $\lambda_1$ and $\lambda_2$ accordingly.
Person 1 arrives at cashier 1 at 9:00.
Person 2 arrives at cashier 2 at 9:01.
Person 3 arrives at 9:03 with both cashiers still being occupied, and flips a fair coin by which he decides to wait for cashier 1 or 2 (after choosing a cashier, he is only waiting for that one).
If it is known that at 9:06, his turn still hasn't arrived yet, what is the probability that at 9:08 it still wouldn't be his turn?
I tried to solve it this way:
Y = overall waiting time for person 3
X = 1 if cashier 1 was chosen, 0 else
$P(Y > 3+2 | Y > 3) = P(Y > 2) = P(Y > 2 | X = 1) \cdot P(X = 1) + P(Y > 2 | X = 0) \cdot P(X = 0 ) = e^{-2\lambda_1} \cdot 0.5 + e^{-2\lambda_2} \cdot 0.5$
Apparently my solution is wrong but I don't understand why. If you could please point to where I'm mistaken and how could it be solved correctly that would be great.
 A: When you write $P(Y>3+2 \mid Y>3) = P(Y>2)$, you assume that $Y$ is a memoryless distribution: an exponential. But $Y$ is not an exponential: $Y$ is the mixture of two exponentials with different rates! So we can't make this simplification yet.
Instead, consider the following four cases:

*

*$X=1$ and $3 < Y < 5$: this happens with probability $\frac12(e^{-3\lambda_1} - e^{-5\lambda_1})$.

*$X=1$ and $Y > 5$: this happens with probability $\frac12 e^{-5 \lambda_1}$.

*$X=2$ and $3 < Y < 5$: this happens with probability $\frac12(e^{-3\lambda_2} - e^{-5\lambda_2})$.

*$X=2$ and $Y > 5$: this happens with probability $\frac12 e^{-5 \lambda_2}$.

These are the probabilities before we are told that $Y>3$. But when we are given that $Y>3$, we know that exactly one of these four cases happens. So we should scale up these four probabilities so that they add up to $1$. They become, respectively:
$$
   \frac{e^{-3\lambda_1} - e^{-5\lambda_1}}{e^{-3\lambda_1} + e^{-3\lambda_2}},\;
   \frac{e^{-5\lambda_1}}{e^{-3\lambda_1} + e^{-3\lambda_2}},\;
   \frac{e^{-3\lambda_2} - e^{-5\lambda_2}}{e^{-3\lambda_1} + e^{-3\lambda_2}},\;
   \frac{e^{-5\lambda_2}}{e^{-3\lambda_1} + e^{-3\lambda_2}}.\;
$$
To get the probability that $Y>5$, we just want to add up the second and fourth case, getting
$$
   \frac{e^{-5\lambda_1} + e^{-5\lambda_2}}{e^{-3\lambda_1} + e^{-3\lambda_2}}.
$$
A: Firstly, I'd like to point out that the exponential distribution is not totally realistic for the time $1$ person takes to check out. It is memoryless, meaning that after a person waits $x$ minutes, the remaining time $y$ they'll need to wait is independent of $x$. $y$ will be the same whether $x$ is $1$ minute or $10$ minutes. So the "If it is known that" clause should be jumping out at you. At the outset, the exponential distribution can be used to forecast the time it will take for something to happen, but once some time has passed, the forecast changes.
If it is known that at 9:06, his turn still hasn't arrived yet, then that is $t=0$ for our equations. The probability that at 9:08 it still wouldn't be his turn is the inverse of the probability that it would be, which is the $\text{OR}$ junction of the two exponential distributions. (The $\text{OR}$ of two probabilities is their sum minus the probability of both occuring.)
$$Pr(X \leq 2 \text{ min}) = 1- e^{-(2/(\lambda_1 \text{in minutes}))} + 1- e^{-(2/(\lambda_2 \text{in minutes}))} - [(1- e^{-(2/(\lambda_1 \text{in minutes}))}) \cdot (1- e^{-(2/(\lambda_2 \text{in minutes}))})]$$
