Respuesta :
Answer:
a) The probability that an article of 10 pages contains 0 typographical errors is 0.8187.
b) The probability that an article of 10 pages contains 2 or more typographical errors is 0.0175.
Step-by-step explanation:
Given : The expected number of typographical errors on a page of a certain magazine is 0.2.
To find : What is the probability that an article of 10 pages contains
(a) 0 and (b) 2 or more typographical errors?
Solution :
Applying Poisson distribution,
[tex]N\sim Pois(0.2)[/tex]
[tex]P(N=r)=\frac{e^{-np}(np)^r}{r!}[/tex]
where, n is the number of words in a page
and p is the probability of every word with typographical errors.
Here, n=10 and E(N)=np=0.2
a) The probability that an article of 10 pages contains 0 typographical errors.
Substitute r=0 in formula,
[tex]P(N=0)=\frac{e^{-0.2}(0.2)^0}{0!}[/tex]
[tex]P(N=0)=\frac{e^{-0.2}}{1}[/tex]
[tex]P(N=0)=e^{-0.2}[/tex]
[tex]P(N=0)=0.8187[/tex]
The probability that an article of 10 pages contains 0 typographical errors is 0.8187.
b) The probability that an article of 10 pages contains 2 or more typographical errors.
Substitute [tex]r\geq 2[/tex] in formula,
[tex]P(N\geq 2)=1-P(N<2)[/tex]
[tex]P(N\geq 2)=1-[P(N=0)+P(N=1)][/tex]
[tex]P(N\geq 2)=1-[\frac{e^{-0.2}(0.2)^0}{0!}+\frac{e^{-0.2}(0.2)^1}{1!}][/tex]
[tex]P(N\geq 2)=1-[e^{-0.2}+e^{-0.2}(0.2)][/tex]
[tex]P(N\geq 2)=1-[0.8187+0.1637][/tex]
[tex]P(N\geq 2)=1-0.9825[/tex]
[tex]P(N\geq 2)=0.0175[/tex]
The probability that an article of 10 pages contains 2 or more typographical errors is 0.0175.
