cookie problem ; solution by Poisson approach doesn't matches binomial one

Question

here's the problem.

A company produces chocolate chip cookies. number of piece of choco chip per one cookie is 2.5 (in Poisson average). These cookies are boxed with 20 pieces together. what's the possibility, that every 20 cookies contain at least one chocolate chip each?

This is my solution

p = possibility that there's no choco chip in one cookie = 1-(e^-2.5) =0.9179

the number of cookies in the box(20 pieces) containing at least one chip is in the binomial distribution of

B(20, 0.9179)

And from this stage, there're two different way to solve this, and both pull out different numbers, unluckily.

1) binomial distribution approach

p2 =possibility that there's no cookie without chip

= 20combination20 * (0.9179)^20

= 0.180322

this seems to be the right answer.

2) Poisson approach B(20, 0.9179)
from this, the Poisson average of the number of cookies containing at least one chip is

20*09179 = 18.3583 = λ

then

possibility every 20 cookie has at least one chip

= P(x=20)

=λ^20 * e^λ / 20!

=0.082721090

the solution 1 and 2 put out totally different numbers, though both seem right approach. Why is it? Thank you very much!

Answer 1

I think

• the probability there is no chip in a particular cookie is $e^{-2.5}\approx 0.082085$
• the probability there is at least one chip in a particular cookie is $1-e^{-2.5}\approx 0.917915$
• the probability there is at least one chip in every cookie is $(1-e^{-2.5})^{20}\approx 0.1803223$ assuming independence

That is essentially your first method, and as you say is correct.

I do not understand your second method. You have calculated what you claim is $P(X=20)$, presumably with $X$ representing the number of cookies with at least one chip. But you could have looked at $P(X=21)$ too and that, on your calculation with $\lambda=18.3583$, would be about $0.0723>0$ even though there are only $20$ cookies. And similarly with $P(X=22)$ etc.