Using the Poisson distribution,
a) The probability that the next page you read contain zero typographical error is 0.81. .
b)The probability that the next page you read contain 2 more typographical errors is 0.017..
Given that expected number of typographical errors on a page out of 10 pages of a certain magazine is , λ = 0.20
Let X is a random variable representing the number of typographical error. We use Poisson distribution. The Poisson distribution formula is ,
P(X) = (e⁻λ λˣ)/x!
For part (a) Put X = 0 , we get
P(X=0) = (0.2)⁰e⁻⁰·²/0! = 1/ e⁰·²⁰= 0.8187
b) Find the probability that the next page you read contains 2 more typographical errors.
That is, P(X≥2) = 1- P(X=0) - P(X=1) = 1 - 0.8187 - (0.2)e⁻⁰·² /1!= 1 - 0.9824= 0.0175
Hence, the probability that an article of 10 pages contains zero typographical error is 0.8187 and contains 2 more error is 0.0175.
To learn more about Poisson distribution, refer:
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Complete question:
6.37. 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 more topography error
