Let X and Y be independent random variables representing the lifetime (in 100 hours) of Type A and Type B light bulbs, respectively.

Both variables have exponential distributions, and the mean of X is 2 and the mean of Y is 3.

Find the joint pdf f(x|y) of X and Y.

Find the conditional pdf f_2(y|x) of Y given X = x.

Find the probability that a Type A bulb lasts at least 300 hours and a Type B bulb lasts at least 400 hours.

Given that a Type B bulb fails at 300 hours, find the probability that a Type A bulb lasts longer than 300 hours.

What is the expected total lifetime of two Type A bulbs and one Type B bulb?

What is the variance of the total lifetime of two Type A bulbs and one Type B bulb?

Let and be independent random variables representing the lifetime (in 100 hours) of Type A and Type B light bulbs, respectively. Both variables have exponential distributions, and the mean of X is 2 and the mean of Y is 3.

a can be solved as follows [tex]\frac{1}{\alpha\beta } e^{\frac{-x-y}{\alpha\beta } }[/tex]

[tex]\frac{1}{6} e^{\frac{-\alpha }{2} -\beta/3 }[/tex]

1/6[tex]e^{\frac{-(3x+2y}{6 }[/tex]

b.

f(y/x)=f(x).f(y)/{f(y)}=f(y)

1/3e^(-y/3)

c. Find the probability that a Type A bulb lasts at least 300 hours and a Type B bulb lasts at least 400 hours.

[tex]\int\limits^\alpha _4 {1/6e^{-(3x+2y)/6} } \, dy[/tex]

1/2e^-(3x+8)/6

[tex]\int\limits^\alpha _3 {e^{-(3x+800)} } \, dx \\[/tex]

0.05882

Given that a type B fails at 300 hours . we find the probability that type A bulb lasts longer than 300hr

f(x>y|y=3)=f(x>3)

[tex]\int\limits^a_3 {1/2e^{-x/2} } \, dx[/tex]

0.2231

e.e) What is the expected total lifetime of two Type A bulbs and one Type B bulb?

E(2A+B)

(2E(A)+E(y)

2*2+3

=7

f. variance of the total lifetime of two types A bulbs and one type B bulb

V(2x+y)=

4*4+9

=25