# 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?

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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!!

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The exponential distribution has a closed form cdf. So it is very easy to do calculations like this. Well done anyway +1 –  Michael Chernick Sep 20 '12 at 0:58
Hi @Michael! I didn't mean to convey sarcasm or any kind of disrespect. I only thought that you may have refrained from showing that $P(Y>5|Y>2) = P(Y>3)$ because you weren't sure how to sure how to proceed with the proof. It's possible that you haven't covered that yet in your PhD studies. When you do, you can always "@" me here if you have any questions! –  user41583 Sep 19 '12 at 23:58