Cfrgtkky

The amount of time needed to wash a car at a car-washing station is
exponentially distributed with an expected value of $15$ minutes. You
arrive at the car-washing station while it is occupied and one other
car is waiting for a wash. The owner of this car informs you that the
car in the washing station has already been there for $10$ minutes.
What is the probability that the car washing station will need no more
than $5$ minutes extra? What is the probability that you have to wait
more than $20$ minutes before your car can be washed?

Try

Let $T$ be time needed to wash a car which is exponential with parameter $lambda = frac{1}{15}$. For the first situation we want

$$ P(T < 15 mid T > 10 ) = dfrac{ P(10 < T < 15 ) }{P(T > 10 )} = dfrac{ int_{10}^{15} 1/15 e^{-t/15} dt }{e^{-10/15} }$$

For the second part, we want $P(T > 20 mid T > 10 )$ . HEre we can apply the memoryless property of the exponential:

$$ P(T>20 mid T>10) = P(T > 10) = e^{-10/15} $$

Is this a correct solution?

HINT 1:
For the first part, I don't see why you didn't also use the memoryless property, since
$$P(T<15|T>10)=1-P(T>15|T>10)=1-P(T>5).$$

HINT 2:
For the second part, note that the total time you'll have to wait is
$$T=T_1+T_2,$$
where $T_1$ is the aditional time that the car which is already in the car-wash will need to finish and $T_2$ is the time it will take to the second car. Also use the memoryless property to determine the distribution of $T_1$ and deduce the distribution of $T$ (you'll have to assume that $T_1$ and $T_2$ are independent, though).

EDIT: Remember the following property:

If $Xsim Gamma(alpha_X,lambda)$, $Ysim Gamma(alpha_Y,lambda)$ and they are independent r.v., then$$X+Ysim Gamma(alpha_X+alpha_Y,lambda).$$

And also remember that $mathcal E(lambda)=Gamma(1,lambda)$.