Probability of transistor survival with exponential MTBF distribution

[math4everyone]

Homework Statement

Homework Equations

$$f_X(x)=\lambda e^{-\lambda x}$$
$$F_X(x) = 1-e^{-\lambda x}$$
$$\mu = \frac{1}{\lambda}$$

The Attempt at a Solution

a) $$f_{X,Y}(x,y) = f_X(x)f_Y(y) = \frac{1}{800} e^{-\frac{1}{800}x} \frac{1}{1000}e^{-\frac{1}{1000}y}$$
$$=\frac{1}{800000}e^{-\left(\frac{1}{800}x+\frac{1}{1000}y\right)}$$
$$F_{X,Y}(x,y) = F_X(x)F_Y(y)=(1-e^{-\frac{1}{800}x})(1-e^{-\frac{1}{1000}y})$$
$$F_{X,Y}(1000,1000) = (1-e^{-\frac{1000}{800}})(1-e^{-\frac{1000}{1000}})=0.451$$
b) I am having trouble with this
c) Apply the same concept of a) and get:
$$F_{X,Y}(1000,1000,1000) = (1-e^{-\frac{1000}{800}})(1-e^{-\frac{1000}{1000}})(1-e^{-\frac{1000}{500}})=0.389$$
d) I am also struggling with this
$$P(\text{won't fail in the next 1000 hours}|\text{didn't fail in the first 1000 hours})$$
$$=\frac{P(\text{won't fail in the next 1000 hours and didn't fail in the first 1000 hours})}{P(\text{didn't fail the first 1000 hours})}$$
$$=\frac{P(\text{won't fail in the next 1000 hours and didn't fail in the first 1000 hours})}{1-F_X(1000,1000,1000)}$$
$$=\frac{P(\text{won't fail in the next 1000 hours and didn't fail in the first 1000 hours})}{0.61}$$
I am having trouble of finding the probability of the numerator... Is it $$(1-F_X(1000,1000,1000))^2$$?

 

[haruspex]

b) I am having trouble with this

Suppose the first fails after time t. What is the probability density for that? How long does the second one have for both to have failed in 1000h? What is the probability density of the overall event?

 

[StoneTemplePython]

If you are familiar with Poisson counting processes -- I'd suggest using that for B -- You basically want the complement to ___.

Using exponential CDFs and their complements should be get you there for the other problems.

Hint: recall that that exponential distributions exhibit memorylessness. This should be immensely helpful in one case and is important to remember in general.