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
[tex]P(X=x)=\frac{e^{-m}m^{x}}{x!}[/tex]
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 :
[tex]P(X=x)=\frac{e^{-0.3}0.3^{x}}{x!}[/tex]
is the probability function.
For a)
[tex]P(X=1)=\frac{e^{-0.3}0.3^{1}}{1!}=0.222[/tex]
For b)
[tex]P(X\geq 2)=1-P(X<2)[/tex]
[tex]P(X\geq 2)=1-[P(X=0)+P(X=1)][/tex]
[tex]P(X\geq 2)=1-(e^{-0.3}+0.222)\\P(X\geq 2)=0.037[/tex]
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)
[tex]P(X=0)=e^{-0.3}[/tex] ⇒
[tex]P(A1).P(A2)=[P(X=0)].[P(X=0)][/tex]
[tex][P(X=0)].[P(X=0)]=(e^{-0.3})(e^{-0.3})=0.549[/tex]
