Question

Asgn: Exponetial pdf's Page 190, #'s 91, 93 #A. Car batteries fail at random wrt milages with an average failure rate of 1 in 10 kilometers. Define X as the number of failures in d kilometers, M as distance in kilometers between failures. Give the probability pdf of X, and the pdf of M, give the mean values of both ranom variables and find the probabilities that: (a) at least one battery fails within 3 km, (b) exactly 3 batteries fail within 5 km, (c) given that there is a failure at 2 km, find the probability that the next failure occurs within the next 5 km

Answer 1

lambda = Average number of batteries in 10 kilometer = 1

a) Number of batteries fails within 3 kilometer = 3/10.

X_d : Number of failures in 3 kilometer

The probability distribution of X_d is Poisson with parameter lambda = 3/10. = 0.30

X_d ~ Poisson ( lambda = 0.30)

P ( at least one battery fails within 3 km) = P ( X_d >= 1) = 1 - P ( X_d = 0)

0.30.30 e 1 -

= 1 - 0.7408

= 0.2592.

b) Number of batteries fails within 5 kilometer = 5/10.

X_d : Number of failures in 5 kilometer

The probability distribution of X_d is Poisson with parameter lambda = 5/10. = 0.50

X_d ~ Poisson ( lambda = 0.50)

P ( Exactly 3 batteries fails within 5 km) = P ( X_d = 3) = 1 - P ( X_d = 0)

0.50.5 e 3!

=0.0126

c) theta = Average failure rate = 1/10.

M : distance in kilometers between failure .

The probability distribution of M is exponential with mean = 1/10

The p.d.f . of M is

f (m) em/10 10 m 0

and cumulative density function

P ( M<= m) = 1 - e^-m/10.

P (M > m) = e^{-m /10}.

Required Probability = P ( M > 7 / M >2)

By using memory less property of an exponential distribution

P ( M > a+ b / M >a) = P ( M >b).

Hence P ( M > 7 / M >2) = P ( M > 5)

= e^-0.5

= 0.6065