- #1

math4everyone

- 15

- 0

## 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 $$(1-F_X(1000,1000,1000))^2$$?