I have a question about a light bulb. It's lifetime is exponentially distributed, with an average of 10,000 hours. a.) Calculate the probability that a randomly chosen lightbulb lasts for more than 20,000 hours?
Here I saw 2 ways to do this, and am not sure which is correct. My first result was: exp(-20000/10000) = exp(-2/1) = 0.135
And my second result: P(X>20000) = 1-P(X≤20000) = 1-(1-exp(-20000*10000)) = exp(-20000*10000) = 0
Which of these is the correct way for this? Also, the next 2 questions were:
(b) What is the probability that a randomly chosen lightbulb lasts for more than 8,000 hours?
(c) Given that a lightbulb has survived for 8,000 hours already, what is the probability it will survive past 20,000 hours?
For (b), would I just do the same as in (a)? And for (c), because exponential distribution are memoryless, I would do the same but for X>12000?
Thanks in advance!
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You are on the right track, remember that if $X$ is distributed exponentially with parameter $\lambda$, then its PDF is
$$ f_X(x) = \lambda e^{-\lambda x} \tag{1} $$
and CDF is
$$ F_X(x) = 1 - e^{-\lambda x} \tag{2} $$
The mean is
$$ \langle X \rangle = \color{blue}{\frac{1}{\lambda}} \tag{3} $$
And there is where your problem is
$$ \langle X \rangle = 10,000~{\rm hours} = \frac{1}{\lambda} ~~~\Rightarrow \lambda = \frac{1}{10,000~{\rm hours}}\tag{4} $$
So that, the probability that a randomly chosen lightbulb lasts for more than $20,000~{\rm hours}$ is just
\begin{eqnarray} P(X > 20,000~{\rm hours}) &=& 1 - F_X(20,0000~{\rm hours}) = 1-\left(1- \exp\left(-\frac{20,000~\cancel{\rm hours}}{10,000~\cancel{\rm hours}}\right)\right) \\ &=& e^{-2/1} = e^{-2} = 0.136 \tag{5} \end{eqnarray}