Player 1 scored 10 points in a hockey game. Player 2 has the most points of any hockey player in history, averaging 2.629 points/game in the 394 regular season games played in 5 seasons '81-'82 to '85-'86. Computer the probability that Player 2 had at least 10 points in at least one of the 394 games.
Textbook Solution:
Poisson Distribution Formula: $$ {\rm P}(Y \geq 10) = \sum\limits_{y = 10}^\infty {\frac{{e^{ - \mu } \mu^y }}{{y!}}} =1-\sum\limits_{y = 0}^9 {\frac{{e^{ - \mu } \mu^y }}{{y!}}} = 1-0.9996=0.0004. $$ Binomial Distribution Formula: $$ {\rm P}(X \geq 1) = 1-{\rm P}(X=0) = 1-C^{394}_0 (.0004)^0 (1-.0004)^{394} = 1-0.8542=0.1458. $$
The first step is to use poisson distribution formula since the mean is given to find at least 10 points in a game in one of the 394 games played (space).
I don't understand why we need to further use the binomial distribution formula when the probability of one game having at least 10 points is 0.0004?
Also...how do I calculate big factorials for the binomial formula? I seem to get an error. I looked it up and I think Stirling’s formula can be used. This isn't covered in the course content and unsure if there is an easier way to calculate big factorials...
The question is asking for the probability that Player 2 has at least one game in which he scored at least $10$ points, among the $394$ games played. If you knew the probability that he would score at least $10$ points for any single game, then it's clear that the desired probability would be found through a binomial distribution calculation, just like if I asked you, if you roll a die $394$ times, what is the probability you would obtain at least one six--you'd need to know the probability of obtaining a six for a single roll.
But you don't know the single-game probability of Player 2 scoring at least $10$ goals; instead, you are given the average goals per game, which is $2.629$. That means you have to use another model, one that applies to a single game's performance. To this end, we treat this number as an average rate of goals per game, and the random number of goals is modeled by a Poisson distribution, thus letting us compute the single-game probability of scoring at least $10$ goals.
Regarding your second question, you do not actually need to compute such a large factorial. Since $\Pr[X = 0] = \binom{394}{0} (0.0004)^0 (1 - 0.0004)^{394 - 0}$, we can see that $$\binom{394}{0} = \frac{394!}{0! (394-0)!} = \frac{394!}{394!} = 1.$$ It's only when $n$ is large and $x$ (and $n-x$) are not small that the binomial coefficient $\binom{n}{x}$ is difficult to calculate exactly; e.g., $\binom{394}{2} = \frac{(394)(393)}{2}$ is still easy, but $\binom{394}{157}$ is not. In such a case, Stirling's approximation is appropriate.