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 μ = 0.4. (Round your answers to three decimal places.)

(a) What is the probability that a disk has exactly one missing pulse?


(b) What is the 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?

Answer 1

According to the given question, 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 μ = 0.4.

Therefore, follows Poisson distribution with the probability mass function,

f (x)-

where

r=0.1.2.3.

(a) Therefore the probability that a disk has exactly one missing pulse is determined as:

P(x = 1) =

e-0.4 × 0.41 1!

0.2681

The required probability that a disk has exactly one missing pulse

0.2681

(b) Therefore the probability that a disk has at least two missing pulses:

$P(x\geq 2)=1-\left [ P(x=0) +P(x=1)\right ]$

0! 1!

єー0.4 × 0.40 0! є_0.4 × 0.41 1!

= 1-10.6703 + 0.26811

$=1-0.9384$

0.06157

$=0.0616$

Therefore the required probability that a disk has at least two missing pulses
0.0616

(c) If two disks are independently selected, therefore the probability that neither contains a missing pulse is determined as:

The probability that the first disk has no missing pulse as

$P(x=0)=\frac{e^{-\mu }\mu ^{0}}{0!}$

$=\frac{e^{-0.4 }\times 0.4 ^{0}}{0!}$

$=0.6703$

The probability that the second disk also has no missing pulse as

$P(x=0)=\frac{e^{-\mu }\mu ^{0}}{0!}$

$=\frac{e^{-0.4 }\times 0.4 ^{0}}{0!}$

$=0.6703$

Therefore as the  two disks are independently selected, therefore the probability that neither contains a missing pulse , is determined as:

0,6703 x0,6703

  $=0.4493$

Therefore the required probability that the probability that neither contains a missing pulse is

0.4493

Answer 2