Alice and Beatrix each buy the same kind of laptop on the same day. The working

Question

Alice and Beatrix each buy the same kind of laptop on the same day. The working life of this type of laptop is an exponential variable with a mean of 10 months. Let A be the working life, in months, of Alice's laptop. Let B be the working life, in months, of Beatrix's laptop. Let X be the amount of time, in months, that both laptops function. Let Y be the amount of time, in months, that at least one

laptop functions.

a) What is the probability that both laptops are still working after x months?

b) Write down the pdf for X. Is X exponentially distributed?

c) What is the probability that at least one laptop is still working after y months?

d) Write down the pdf for Y. Is Y exponentially distributed?

e) Compute E[X] and E[Y]. (Hint: the "best" solution for this part doesn't involve any integrals.)..

Explanation / Answer

As the mean working life is 10 months, and the mean of the exponential is 1/lambda, 1/lambda = 10, so lambda = 1/10

The survival function for each laptop is e^(-lambda x)

a)P(both are working after x month) = P(the first one is still working)*P(second one is still working) =

e^(-1/10 x) * e^(-1/10 x) = e^ (-1/10 x + -1/10 x) = e^ - 2/10 x = e^ -1/5 x

b)The pdf is - the derivative of the survival function = - -1/5 e^ -1/5 x = 1/5 e^ -1/5 x for x >= 0. Yes, X is exponentially distributed with parameter lambda = 1/5

c)P(at least one is still working after y months) = 1 - (p both fail) = 1 - (1 - e^ -1/10 y)^2 =

1 - (1 - 2e^-1/10 y + e^ -1/5 y) =

2e^-1/10 y - e^ -1/5 y

d) The pdf is - the derivative of the survival function is

2/10e^-1/10y - 1/5e^-1/5y =

1/5e^-1/10y - 1/5e^-1/5y for y >= 0 This is not an exponential distribution

e) For E[X], this is the exponential distribution with parameter 1/5, so the mean is 1/(1/5) = 5

For E(y), note that 1/5e^-1/10y - 1/5e^-1/5y = 2(the exponential distribution with parameter 1/10) - the exponential distribution with parameter 1/5

Thus, the mean is 2(1/(1/10) - 1/(1/5) = 2*10 - 5 = 20 - 5 = 15

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