Question

Assume the number of errors on each page of a book are independent,
and the number of errors on a particular page can be described by a Poisson random variable
with parameter 1/2
. If the book is 786 pages long, what is the probability that there are more
than 50 pages with at least one error on it? Hint: Whether or not a page has an error can
be viewed as a Bernoulli trial.

Answer 1

What's the probability that a page has NO error on it? Poisson will tell you.

Answer 2

okay let me try that

Answer 3

still confused

Answer 4

As the previous user hinted, the probability that an error occurs on any given page is \(\dfrac{1}{2}\), the same as the parameter for the Poisson distribution.

If \(X\) is the number of pages with at least one error on it, then you're trying to find \(\mathbb P(X>50)=1-\mathbb P(X\le50)=\displaystyle\sum_{x=0}^{50}p(x)\) where \(p(x)\) is the probability of getting \(x\) erroneous pages. Are you familiar with the density function for a binomial distribution?

Answer 5

yes i understand that part but how am i supposed to get the probability for P(X≤50)?

Answer 6

i know the equation for P(X≤50) = 1-(1-p)^50

Answer 7

but how do i solve for P? would i have to use Poisson?

Answer 8

That "success" probability \(p\) is \(\dfrac{1}{2}\), which is what the Poisson distribution tells you.

Answer 9

oh so i dont have to use the poisson formula?

Answer 10

I always wonder why you are given problems with no way to solve them.

\(p(0) = \dfrac{\lambda^{0}e^{-\lambda}}{0!}\)

\(p(1) = \dfrac{\lambda^{1}e^{-\lambda}}{1!}\)

\(p(2) = \dfrac{\lambda^{2}e^{-\lambda}}{2!}\)

...

\(p(n) = p(n-1)\cdot\dfrac{\lambda}{n}\)

Answer 11

so i would have to do this all the way up to 50?

Answer 12

Absolutely not. Use Normal Approximation. p(at least one error) = 1 - p(0 error)

786 pages

mean = 786 * 0.60653066 = 476.7330985 = Expected # of pages with NO error.
Standard Deviation = sqrt[786 * 0.60653066 * (1- 0.60653066)] = 13.69597961 = SD of pages with NO error.

Z-score for at least 50 pages with at least one error: \(\dfrac{(786 - 50)-476}{13.69597961} = 18.9836\)

A z-score of 19ish? Forget it. You're going to get at least 50 pages with at least 1 error.

Think it through carefully. Understand each piece.

Answer 13

oh okay i see. i used normal approximation earlier but got like -10