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

a) 0.222

b)0.037

c)0.549

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)=(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)=(e^(-0.3)0.3^(x))/(x!)`

is the probability function.

For a)

`P(X=1)=(e^(-0.3)0.3^(1))/(1!)=0.222`

For b)

`P(X\geq 2)=1-P(X<2)`

`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`