Question

A device is used in many kinds of systems. Assume that all systems have either 1, 2, 3, or 4 of these devices and that each of these four possibilties is equally likely to be the case. Each device in a system has probablility = 0.1 of failing, and the devices function independently of one another. This implies that once we know how many devices are present, the probability distribution of the number of failures will be known. E.g. if a system employs 3 of the devices, then the number that fail will have a binomial distribution with parameters n = 3 and p = 0.1

Denote with X, the number of failures of devices in the system, and with Y, the total number of devices in the system. What is observed is that for b = 1,2,3, and 4, the conditional probability mass function is the binomial probability mass function with parameters n = b, and p = 0.1

a) Find the joint probability mass table of P(X,Y)

Answer 1

P(X | Y = b) ~ Binom(n = b, p = 0.1)

Thus,

P(X = x | Y = b) = ^bC_x * 0.1^x * (1 - 0.1)^b-x = ^bC_x * 0.1^x * 0.9^b-x    for b x

P(X = x | Y = b) = 0 for b < x

Also, P(Y = b) = 0.25 as each of these four possibilties is equally likely to be the case

P(X = x, Y = b) = P(X = x | Y = b) * P(Y = b) = P(X = x | Y = b) * 0.25

As, the number of devices in a system is between 1 and 4. Thus, range of Y is (1,4)

The number of failures can be between 0 and 4. Thus, range of X is (0,4)

The joint probability mass table of P(X,Y) is,

	Y = 1	Y = 2	Y = 3	Y = 4
X = 0	1C0 * 0.10 * 0.91 * 0.25 = 0.225	2C0 * 0.10 * 0.92 * 0.25 = 0.2025	3C0 * 0.10 * 0.93 * 0.25 = 0.18225	4C0 * 0.10 * 0.94 * 0.25 = 0.164025
X = 1	1C1 * 0.11 * 0.90 * 0.25 = 0.025	2C1 * 0.11 * 0.91 * 0.25 = 0.045	3C1 * 0.11 * 0.92 * 0.25= 0.06075	4C1 * 0.11 * 0.93 * 0.25 = 0.0729
X = 2	0	2C2 * 0.12 * 0.90 * 0.25 = 0.0025	3C2 * 0.12 * 0.91 * 0.25 = 0.00675	4C2 * 0.12 * 0.92 * 0.25 = 0.01215
X = 3	0	0	3C3 * 0.13 * 0.90 * 0.25 = 0.00025	4C3 * 0.13 * 0.91 * 0.25 = 0.0009
X = 4	0	0	0	4C4 * 0.14 * 0.90 * 0.25 = 0.000025