Conditional probabilities involving the exponential distribution The number of years the laptop functions is exponentially distributed with mean = 5 years. If a customer purchased an old laptop which was used for last two years, what is the probability that it will function for at least 3 years?
 A: The probability you want is $P(X>5|X>2)$ where $X$ has an exponential distribution. We're seeing whether it's greater than 5 because you want to know if it lasts 3 additional years, after it's already been functioning for 2 years. 
Using the CDF of the exponential distribution, with $\lambda$ referring to the "rate parameter", 
$$ P(X \leq 5 | X> 2 ) = \frac{ P( X \leq 5 \ \cap \ X > 2) }{P(X>2)} = \frac{ P( 2 < X \leq 5 ) }{P(X>2)} = \frac{ e^{-2\lambda} - e^{-5 \lambda} }{e^{-2 \lambda} } = 1 - e^{-3\lambda}$$
This means $$P(X>5|X>2) = 1- P(X \leq 5 | X> 2 ) = e^{-3 \lambda}$$
In your case the mean is 5, which means $\lambda = 1/5$ using the parameterization I've used here, so the probability you want is $e^{-3/5} \approx .549$. 
Note that this is the same as the probability that a brand new computer lasts for 3 years. This is because the exponential distribution has something called the memoryless property. 
I hope this helps!!
A: Based on the lack of memory property for the exponential distribution, the remaining life has the same distribution as it would have if the laptop were brand new.  So just integrate the given exponential density from 3 to infinity to obtain the desired probability since P[X>5|X>2]=P[X>3].
A: Because means is 5 we have that $\lambda = \frac{1}{5}$ and you need $P(X>5|X>2).$
$$\begin{align}
P(X>5|X>2) &= \frac{P(X>5 \cap X > 2)}{P(X>2)}\\
 &= \frac{P(X>5)}{P(X>2)}\\
 &= \frac{\int_5^{\infty}\frac{1}{5}e^{-\frac{1}{5}x}}{\int_2^{\infty}\frac{1}{5}e^{-\frac{1}{5}x}}\\
 &= e^{-1}/e^{-\frac{2}{5}}\\
 &=e^{-\frac{3}{5}}
\end{align}$$
