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 = 0.3. (Round your answers to three decimal places.) (a) What is the probability that a disk has exactly one missing pulse? .222 Correct: Your answer is correct. (b) What is the probability that a disk has at least two missing pulses? .037 Correct: Your answer is correct. (c) If two disks are independently selected, what is the probability that neither contains a missing pulse? .122 Incorrect: Your answer is incorrect.

Answer 1

Answer:

a) 0.222

b)0.037

c)0.549

Step-by-step explanation:

Let's start defining the random variable ⇒

X : ''The number of missing pulses''

X can be modeled as a Poisson random variable.

X ~ Po(λ)

In a Poisson distribution : μ = λ

Where μ is the mean of the variable.

X ~ Po(0.3)

The probability function for a Poisson random variable is :

In the equation I replace λ = m

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

Where P(X=x) is the probability of the random variable X to assume the value x

e is the euler number

m = λ is the mean of the variable

In this exercise :

$P(X=x)=\frac{e^{-0.3}0.3^{x}}{x!}$

is the probability function.

For a)

$P(X=1)=\frac{e^{-0.3}0.3^{1}}{1!}=0.222$

For b)

$P(X\geq 2)=1-P(X$

$P(X\geq 2)=1-[P(X=0)+P(X=1)]$

$P(X\geq 2)=1-(e^{-0.3}+0.222)\\P(X\geq 2)=0.037$

c)

Let's define A :''a disk doesn't contain a missing pulse''

We are looking for P(A1∩A2) of two different disk don't have a missing pulse.

Because of the independence we can write this probability as

P(A1∩A2)= P(A1).P(A2)

The probability of a random disk to don't have a missing pulse is P(X=0)

$P(X=0)=e^{-0.3}$ ⇒

$P(A1).P(A2)=[P(X=0)].[P(X=0)]$

$[P(X=0)].[P(X=0)]=(e^{-0.3})(e^{-0.3})=0.549$