$\lambda$= 1/3 (one failure every 3000 hours, on average). i) Determine the probability that lamp will last longer than its mean life of 3000 hours. ii) Determine the probability that the lamp will last between 2000 and 3000 hours. iii) Find the probability that the lamp will last for another 1000 hours, given that it is operating after 2500 hours.

i) Probability the lamp will last longer than it's mean life

$\begin{aligned} &=p(x\gt3) \\ &=1- p( x \leq 3) \\ &=1-\left(1-e^{-3 / 3}\right) \\ &=e^{-1} \\ &= 0.368 \end{aligned}$

ii) Probability the industrial lamp will last between 2000 and 3000 hours is

$\begin{aligned} &=P(2 \leq x \leq 3) \\ &=F(3)-F(2) \\ &=\left(1-e^{-\frac{3}{3}}\right)-\left(1-e^{-2 / 3}\right) \\ &=\left(1-e^{-1}\right)-\left(1-e^{-2 / 3}\right) \\ &=0.632-0.487 \\ &= 0.145 \end{aligned}$

iii) Probability that the lamp will last for another 1000 hours given that it is operating after 2500 hours

$\begin{aligned} &=p(x\gt3.5 | x\gt2.5) \\ &=p(x\gt2.5+1 / x\gt2.5) \\ &=p(x\gt1) \\ &=P-P(x \leq 1) \\ &=1-\left(1-e^{-1 / 3}\right) \\ &=e^{-1 / 3} \\ &=0.717 \end{aligned}$