Question

Suppose 300 misprints are distributed randomly throughout a book of 500 pages. Find the probability that a given page contains (1) exactly 2 misprints, (2) 2 or more misprints, (3) nomisprints, and (4) between 3 and 5 misprints inclusive.​(answer with explanation)

Answer 1

Answer:

Given that,

Misprints = 300

Pages of book = 500

We need to calculate the value of λ

Using formula of λ

$\lambda=\dfrac{number\ of\ misprints}{number\ of\ pages}$

Put the value into the formula

$\lambda=\dfrac{300}{500}$

$\lambda=0.6$

We need to calculate the probability

Using formula of probability

$P(x)=\dfrac{e^{-\lambda}\times(\lambda)^{x}}{x!}$

For x=4,

Put the value of x

$P(4)=\dfrac{e^{-0.6}\times(0.6)^{4}}{4!}$

$P(4)=0.0029$

Hence, The probability is 0.0029.

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Answer 2

Answer:

Misprints = 300

Pages of book = 500

We need to calculate the value of λ

Using formula of λ

\lambda=\dfrac{number\ of\ misprints}{number\ of\ pages}λ=

number of pages

number of misprints

Put the value into the formula

\lambda=\dfrac{300}{500}λ=

500

300

\lambda=0.6λ=0.6

We need to calculate the probability

Using formula of probability

P(x)=\dfrac{e^{-\lambda}\times(\lambda)^{x}}{x!}P(x)=

x!

e

−λ

×(λ)

x

For x=4,

Put the value of x

P(4)=\dfrac{e^{-0.6}\times(0.6)^{4}}{4!}P(4)=

4!

e

−0.6

×(0.6)

4

P(4)=0.0029P(4)=0.0029

Hence, The probability is 0.0029.