Probabillity of failures involving exponential distribution

I have come across this problem:

A room is lit by $10$ light bulbs. The lifetime, X, of the light bulbs follows an exponential distribution with mean $\mu$ = $1000$ hours. In a time window of $2000$ hours:

1. what is the expected number of failures?

I did this as: For time window of 2000 hours, and 10 bulbs, expected value= $\frac{1}{1000}$ * $2000$ * $10$ = $20$

1. what is the probability that there will be more than $5$ failures?

The light bulbs are replaced immediately upon failure.

I calculated the probability that a light bulb will not fail (X> $2000$) is:

$\mu$ =$1000$ , $\lambda$ = $\frac{1}{1000}$

P(X>2000) = $e^{-\lambda x}$ = $e^{-2000/1000}$ = 0.13 ( 13% )

and tried to incorporate that in to solving the questions but I think I'm heading the wrong way.
Edit: I also used Cumulative Poisson Probability to find upper/lower limits but I don't know how to apply it to multiple objects. Also I am curious as to how the problem would change if the light bulbs were not being replaced upon failure.

• You need to use the poisson distribution with parameter $\lambda=2000/\mu$ to answer the more involved questions. – kodlu Nov 17 '15 at 9:06
• So let me see if i get this. I will use P(x; μ) = (e-μ) (μx) / x! whith $\lambda$ instead of $\mu$ and multiply the results by 10 for the number of lught bulbs? – apot Nov 17 '15 at 21:32
• Twice I cleaned up your very very clumsy use of MathJax and both times you undid my edits. – Michael Hardy Nov 18 '15 at 4:39
• Yes I am very sorry for that. I did it while adding some stuff to the question and accidentally erased yours. I am very new to this plus the timing was terrible. Doing my best to learn, thanks for your edits. – apot Nov 18 '15 at 4:42

If there were just one light bulb with an average lifetime of $1000$ hours, then the number of failures in time $t$ would have a Poisson distribution with expected value $t/1000$. With $10$ light bulbs functioning independently of each other, that expected value is simply multiplied by $10$. The reason is that the distribution of the sum of independent Poisson-distributed random variables is Poisson-distributed. I think the question of why that is so has been answered here a number of times.