In a book of 500 pages, it is found that there are 250 typing errors. Assume that Poisson law holds for the number of errors per page. Then, the probability that a random sample of 2 pages will contain no error, is :

Question

In a book of 500 pages, it is found that there are 250 typing errors. Assume that Poisson law holds for the number of errors per page. Then, the probability that a random sample of 2 pages will contain no error, is : (A) e-0.3 (B) e-0.5 (C) e-1 (D) e-2

Tardigrade Admin · Accepted Answer

Here number of errors per page =(250/500)=(1/2) and n=2 ∴ λ=n p=2 × (1/2)=1 and probability of no error P(X=0)=(e-1 ×(1)0/0 !)=e-1

Q. In a book of $500$ pages, it is found that there are $250$ typing errors. Assume that Poisson law holds for the number of errors per page. Then, the probability that a random sample of $2$ pages will contain no error, is :

$e^{- 0.3}$

$e^{- 0.5}$

$e^{- 1}$

$e^{- 2}$

Here number of errors per page
$= \frac{250}{500} = \frac{1}{2}$
and $n = 2$
$∴ λ = n p = 2 \times \frac{1}{2} = 1$
and probability of no error
$P (X = 0) = \frac{e ^{- 1} \times ( 1 ) ^{0}}{0 !} = e^{- 1}$

Q. In a book of 500 pages, it is found that there are 250 typing errors. Assume that Poisson law holds for the number of errors per page. Then, the probability that a random sample of 2 pages will contain no error, is :

e−0.3

e−0.5

e−1

e−2