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sergij07 [2.7K]
3 years ago
13

The expected number of typographical errors on a page of a certain magazine is .2. What is the probability that an article of 10

pages contains (a) 0 and (b) 2 or more typographical errors? Explain your reasoning!
Mathematics
1 answer:
Pavel [41]3 years ago
6 0

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,

N\sim Pois(0.2)

P(N=r)=\frac{e^{-np}(np)^r}{r!}

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,

P(N=0)=\frac{e^{-0.2}(0.2)^0}{0!}

P(N=0)=\frac{e^{-0.2}}{1}

P(N=0)=e^{-0.2}

P(N=0)=0.8187

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 r\geq 2 in formula,

P(N\geq 2)=1-P(N

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

P(N\geq 2)=1-[\frac{e^{-0.2}(0.2)^0}{0!}+\frac{e^{-0.2}(0.2)^1}{1!}]

P(N\geq 2)=1-[e^{-0.2}+e^{-0.2}(0.2)]

P(N\geq 2)=1-[0.8187+0.1637]

P(N\geq 2)=1-0.9825

P(N\geq 2)=0.0175

The probability that an article of 10 pages contains 2 or more typographical errors is 0.0175.

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