# calculate the probabilities for a specific Poisson distribution example

i want a solution for this question

The number of goals scored at State College hockey games follows a Poisson distribution with a mean of 3 goals per game. Find the probability that each of four randomly selected State College hockey games resulted in six goals being scored.

i tried this solution but the answer should be 0.00000546

$$\text{mean} = m,\qquad P[x] = \frac{e^{-m} m^x}{x!}$$ $$m = 3,\qquad P[6] = \frac{e^{-3} 3^6}{6!} ≈ 0.0504$$ $$n = 4,\qquad p = 0.0504$$ $$\Pr = 0.0504^4 ≈ 6.4524\times 10^{-6}$$

-
What you did looks fine to me (except perhaps rounding the intermediary result; the probability is approximately $0.645723\cdot10^{-6}$). – David Mitra Mar 31 '12 at 22:31
@Nammari : Like most people you are gullible about what you get from a calculator. If you take $e^{-3} 3^6/6!$ and round it to four digits after the decimal point, then raise to the fourth power, then round again, you get $6.4524\times10^{-6}$. If you wait until the last step to round, you get $6.4572\times 10^{-6}$. The last is correct in as many digits as it shows. To believe the last digits of that first one is silly. People should stop reposing religious faith in calculators, using them as tranquilizing drugs, etc. – Michael Hardy Mar 31 '12 at 22:47
@MichaelHardy i know that but what the book says it 0.00000546 and i cant get this value – Nammari Mar 31 '12 at 22:53
What book is it? – Michael Hardy Mar 31 '12 at 22:57
If the problem is as you posed it, your solution is correct (aside from the rounding issue). The book is wrong; no matter how sloppy I tried to round, I could not get their result. – David Mitra Mar 31 '12 at 23:06