Assuming that the two machines starts at the same time, what is the probability that machine A manufactures the first product?

Question

A company has two types of machines (A and B) to manufacture the same exact product. The manufacturing time of each product is independent of the others and can be modeled as an exponential random variable with λA = 2 items per day for Machine A and λB = 3 items per day for Machine B. Assume the idle time of each machine to be zero.

Assuming that the two machines starts at the same time, what is the probability that machine A manufactures the first product?

Answer 1

Consider X to be the number of products machine A generates in a day, and Y to be the number of products machine B generates in a day.

Basically, we want to compute P(X > Y) [if A makes more products in a day than B, then it must have manufactured the first product since it worked faster than B that day]

Considering the input parameters and the given distributions for each machine, we can obtain the pdf's for x and y:

f(x) = λA*e^-λAx = 2*e^-2x for x >= 0, 0 otherwise

f(y) = λB*e^-λBy = 3*e^-3y for y >= 0, 0 otherwise

Because the manufacturing times are independent, the joint pdf of x and y is the product of the individual pdfs: f(x, y) = f(x)*f(y).

To compute P(X > Y), we evaluate the following the integral of f(x, y):

P(X > Y) = ∫₀^∞∫₀^xf(x, y) dy*dx = ∫₀^∞∫₀^x 6*e^{-2x - 3y} dy*dx = 0.6