Question

The quality of computer disks Is measured by sending the disks through a certifier that counts the number of missing pulses. A certain brand of computer disk has averaged 0.1 missing pulse per disk. Find the probability that two next inspected disk will have no missing pulse, Find the probability that the next inspected disk wkl have more than one missing pulse, Find the probabihty that neither of the next two inspected disks will contain any missing pulse.

Answer 1

Answer:

a) $P(X=0) = 0.1^0 \frac{e^{-0.1}}{0!}=0.905$

b) $P(X\geq 1) = 1-P(X$

c) $P(X=0)=0.2^0 \frac{e^{-0.2}}{0!}=0.819$

Step-by-step explanation:

Definitions and concepts

The Poisson process is useful when we want to analyze the probability of ocurrence of an event in a time specified. The probability distribution for a random variable X following the Poisson distribution is given by:

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

And the parameter $\lambda$ represent the average ocurrence rate per unit of time.

On this case the mean is given by $\lambda =0.1$

Part a:  Find the probability that two next inspected disk will have no missing pulse

For this case we want this probability:

$P(X=0) = 0.1^0 \frac{e^{-0.1}}{0!}=0.905$

Part b: Find the probability that the next inspected disk wkl have more than one missing pulse

For this case we want this probability:

$P(X\geq 1) = 1-P(X$

Part c: Find the probabihty that neither of the next two inspected disks will contain any missing pulse.

For this case the new mean would be $\lambda = 0.1x2 = 0.2$ since we have now two inspected disks, and we want this probability:

$P(X=0)=0.2^0 \frac{e^{-0.2}}{0!}=0.819$