Question

The number N of devices that a technician must try to repair during the course of an arbitrary workday is a random variable having a geometric distribution with parameter p = 1/8. We estimate the probability that he manages to repair a given device to be equal to 0.95, independently from one device to another

a) What is the probability that the technician manages to repair exactly five devices, before his second failure, during a given workday, if we assume that he will receive at least seven out-of-order devices in the course of this particular workday?

b) If, in the course of a given workday, the technician received exactly ten devices for repair, what is the probability that he managed to repair exactly eight of those?

c) Use a Poisson distribution to calculate approximately the probability in part (b).

d) Suppose that exactly eight of the ten devices in part (b) have indeed been repaired. If we take three devices at random and without replacement among the ten that the technician had to repair, what is the probability that the two devices he could not repair are among those?

Answer 1

The probability that he manages to repair a given device to be equal to .

a) The probability that the technician manages to repair 5 devices before his second failure in the 7th repair is (he must repair 5 devices out of 6, fail in the 7th)

b) The number of devices repaired out of has binomial distribution.

The PMF of is  

The probability that he managed to repair exactly eight of those is

c) We assume the Binomial by Poisson distribution with parameter .

The Poisson PMF is

Here . So,

This is only approximate. Exact one is in part(b).

d) Here we have to use Hypergeometric distribution for sampling without replacement .

The number of ways of selecting 3 devices from 10 is .

The number of ways of selecting 2 defective devices from 2 defectives is

The probability that the two devices he could not repair are among those 3 selected is