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)

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

Step-by-step explanation:

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

In which

x is the number of sucesses

e = 2.71828 is the Euler number

is the mean in the given time interval.

In this problem, we have that:

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

This is P(X = 1).

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

We want . So

In which

So

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,

0.7559 = 75.59% probability that neither contains a missing pulse