Question

The run times of two computer programs, A and B, are independent random variables. The run time of program A is an exponential random variable with mean 3 minutes, while that of program B is an Erl(2,lambda) random variable with mean 3 minutes. If both programs are being executed on two identical computers, what is the probability that program B will finish first?

Answer 1

A and B are two ind. RVs, and The run time of program A follows an exp() distribution(say X), whose mean = 1/=3, Thus =1/3

and The run time of program B follows an Erl(2,) distribution (say Y), whose mean = 2/=3, Thus =2/3.

We are to find out the probability that the P(Y>X) P(X<Y) (as both are equivalent) =

Now the DF of a exp distribution with parameter is given as 1-exp(-), Now, = = 1/3 ,

because as it is the integral of a PDF, 3/2=1, and as it is the mean of a exp(1) distribution.

Thus the required probability is 1- = 1-(2/3)² = 1-4/9 = 5/9