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.

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 $$ \frac{e^{-5\lambda_1} + e^{-5\lambda_2}}{e^{-3\lambda_1} + e^{-3\lambda_2}}. $$