Question

For a situation where 100 misprints are distributed randomly throughout the 100 pages of a book, the probability that a page selected at random will contain at most one misprint will be?

Answer 1

Probability that a page selected at random will contain at most one misprint will be 2/e = 0.736

Step-by-step explanation:

100 Misprint in 100 pages

=> λ = 1

the probability that a page selected at random will contain at most one misprint will be = P(0)  + P(1)

=  P(X=x)=  e^(−λ) *  (λ)ˣ / x!

=  e⁻¹  * 1⁰/0!   +  e⁻¹  * 1¹/1!

= e⁻¹ + e⁻¹

= 2e⁻¹

= 2/e

= 2 * 0.368

= 0.736

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Answer 2

The probability that a page selected at random will contain at most one misprint will be 0.736

Step-by-step explanation:

Total number of misprints=100

Total number of pages in a book=100

Expected mean value=$\lambda=\frac{100}{100}=1$

Poisson distribution formula for probability :

$P(X=x)=\frac{e^{-\lambda}\lambda^x}{x!}$

Using poisson distribution formula

$P(x\leq 1)=P(x=0)+P(x=1)$

$P(x\leq 1)=\frac{e^{-1}(1)^{0}}{0!}+\frac{e^{-1}(1)}{1}$

$P(x\leq 1)=e^{-1}+e^{-1}=2e^{-1}$

$P(x\leq 1)=\frac{2}{e}$

$P(x\leq 1)=0.736$

Hence, the probability that a page selected at random will contain at most one misprint will be 0.736.

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