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

Answer:

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:

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

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).

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

$P(X = 1) = \frac{e^{-0.14}*0.14^{1}}{(1)!} = 0.1217$

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

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

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

$P(X = 0) = \frac{e^{-0.14}*(0.14)^{0}}{0!} = 0.8694$

$P(X = 1) = \frac{e^{-0.14}*0.14^{1}}{(1)!} = 0.1217$

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

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