Answer:
1. 0.108368
2. 0.188876
Explanation:
Let X be the exponential random variable that represents the lifetime of a computer.
i.e.
![X\sim Exp(0.001)](https://img.qammunity.org/2021/formulas/mathematics/college/jf6mvlyxou9l4sifdl40vdp6ddde4ajjbi.png)
The probability that the computer will function more than 2000 days can be computed as follows:
P(X > 2000)
:
![f_X(x) = \lambda e^{-\lambda x](https://img.qammunity.org/2021/formulas/mathematics/college/gp8i3nfq7kef5qgn1vf3cd2ei3cph25kxq.png)
P(X > 2000) = 1 - P(X< 2000)
P(X > 2000) = exp(-2000/β) = e⁻²²
P(X > 2000) = 0.108368
2.
By applying conditional probability;
![P(X>2000 \bigg | X>500) =(P(X>2000 \ \cap \ X>500))/(P(X> 500))](https://img.qammunity.org/2021/formulas/mathematics/college/xd3c7tg38fcz1weka9mxtg2giupdg7gk6r.png)
![P(X>2000 \bigg | X>500) =(P(X>2000 ))/(P(X> 500))](https://img.qammunity.org/2021/formulas/mathematics/college/yb5px32vehhuj9tab1nr3b80yl78ap4ct9.png)
![P(X>2000 \bigg | X>500) =(0.108368 )/(0.57375)](https://img.qammunity.org/2021/formulas/mathematics/college/iqel6c3fp04dnk00jcilu3hkmsx97qr4h3.png)
![\mathbfP(X>2000 \bigg](https://img.qammunity.org/2021/formulas/mathematics/college/evfwnx16y9kjeoay30eu94nbaw4nze5mb0.png)