Question

Consider writing onto a computer disk and then sending it through a certifier that counts the number of missing pulses. Suppose this number X has a Poisson distribution with parameter ? = 0.14.

(a) What is the probability that a disk has exactly one missing pulse? (Round to four decimal places)

(b) What is the probability that a disk has at least two missing pulses? (Round to four decimal places)

(c) If two disks are independently selected, what is the probability that neither contains a missing pulse?(Round to four decimal places)

Answer 1

a) 0.1217 = 12.17% probability that a disk has exactly one missing pulse

b) 0.0089 = 0.89% probability that a disk has at least two missing pulses

c) 0.7559 = 75.59% probability that neither contains a missing pulse

Explanation:

In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:

`P(X = x) = (e^(-\mu)*\mu^(x))/((x)!)`

In which

x is the number of sucesses

e = 2.71828 is the Euler number

`\mu` is the mean in the given time interval.

In this problem, we have that:

`\mu = 0.14`

(a) What is the probability that a disk has exactly one missing pulse? (Round to four decimal places)

This is P(X = 1).

`P(X = x) = (e^(-\mu)*\mu^(x))/((x)!)`

`P(X = 1) = (e^(-0.14)*0.14^(1))/((1)!) = 0.1217`

0.1217 = 12.17% probability that a disk has exactly one missing pulse

(b) What is the probability that a disk has at least two missing pulses? (Round to four decimal places)

Either a disk had at most 1 one missing pulse, or it had at least two. The sum of the probabilities of these events is 1. So

`P(X \leq 1) + P(X \geq 2) = 1`

We want
`P(X \geq 2)`. So

`P(X \geq 2) = 1 - P(X \leq 1)`

In which

`P(X \leq 1) = P(X = 0) + P(X = 1)`

`P(X = x) = (e^(-\mu)*\mu^(x))/((x)!)`

`P(X = 0) = (e^(-0.14)*(0.14)^(0))/(0!) = 0.8694`

`P(X = 1) = (e^(-0.14)*0.14^(1))/((1)!) = 0.1217`

`P(X \leq 1) = P(X = 0) + P(X = 1) = 0.8694 + 0.1217 = 0.9911`

So

`P(X \geq 2) = 1 - P(X \leq 1) = 1 - 0.9911 = 0.0089`

0.0089 = 0.89% probability that a disk has at least two missing pulses

(c) If two disks are independently selected, what is the probability that neither contains a missing pulse?(Round to four decimal places)

Each disk has a 0.8694 probability of having no missing pulses.

Since they are independently selected,

`P = (0.8694)^(2) = 0.7559`

0.7559 = 75.59% probability that neither contains a missing pulse