The waiting time (measured in minutes) until the arrival of a bus at a bus station is distributed exponentially such that $X_A \sim Exp(0.1)$ but if the station is busy then the waiting time is also distributed exponentially such that $X_B \sim Exp(0.05)$, the probability that the bus station is busy is $P = 0.18$.
A person randomly arrives at the bus station, it is known that after 8 minutes of waiting he's still standing there, what is the probability that he will have to wait at most 7 more minutes till the bus comes?
My first approach was:
Let's denote in Y the time a person waits for the bus, therefore we need to calculate:
$P(Y\leq 15 | Y \geq 8) = 1 - P(Y > 15|Y\geq 8) =$ [by memorylessness] = $1-P(Y>7) = 1-(0.82\times e^{-7\times0.1} + 0.18\times e^{-7*0.05}) \approx 0.4654$
But when I am looking at the answers they did something like that:
$P(Y\leq 15 | Y \geq 8) = \frac{P(Y\leq15 \cap Y\geq8)}{P(Y\geq8)} = \frac{P(8\leq Y\leq15)}{P(Y\geq8)}= \frac{P(Y\leq15) - P(Y\leq8)}{P(Y\geq8)}=\frac{1-(0.82\times e^{-15\times0.1} + 0.18\times e^{-15*0.05}) - 0.5109}{1-0.5109} \approx 0.4521$
When $P(Y\leq 8) = 0.82\times P(X_A\leq8) + 0.18 \times P(X_B \leq 8) = 0.82(1-e^{-0.1 \times 8}) + 0.18(1-e^{-0.05 \times 8}) = 0.5019$
Can someone shed some light please on why I am getting different answers ? Thanks
The difference arises because the mixture distribution that takes into account "busy" versus "non-busy" states of the station, no longer has the memoryless property. So you cannot claim $\Pr[Y > 15 \mid Y \ge 8] = 1 - \Pr[Y > 7]$ because $Y$ is not exponentially distributed. Only $X_A$ and $X_B$ are exponential.
You can see this more clearly if you remove $X_B$ and simply assume that the arrival time is exponentially distributed as $X_A$, i.e., use both approaches to compute the probability under the assumption that $Y = X_A$. You will get the same result.