# Binomial probability given the mean

You examine each tree in a 25 ft by 25 ft section of forest and record whether or not that tree is infected. Based on historical data, the average number of trees that would be expected to be infected in a section of this size is $3$.

What is the probability that you would find at least $2$ infected trees in a section of this size?

(a) $0.7760$

(b) $0.8009$

(c) $0.2240$

(d) $0.5768$

My impression was that there is insufficient information. If the number of trees in the plot were given, I would compute $$\sum_{k=2}^N \binom Nk \bigg(\frac{3}N\bigg)^k \bigg(1-\frac{3}N\bigg)^{n-k}$$ but I don't have the denominator $N$. Is there a way to solve this?

• This does depend on the number of trees in the plot. – Fimpellizieri Feb 8 '18 at 15:37
• Is there a closed form for the summation above? If so, could I solve for each of the answer choices and check if the $N$ satisfying the equation is plausible? – Tiwa Aina Feb 8 '18 at 15:40
• Well, you can use $1 - \mathbb P(X=0) - \mathbb P(X=1)$ which seems to be simpler. – Fimpellizieri Feb 8 '18 at 15:43

HINT...This requires the use of the Poisson distribution with parameter $\lambda=3$, find $p(X\geq2)$
• $\sum_{x=2}^\infty \frac{3^x}{e^3 x!} = 1 - 4/e^3 \approx 0.8009$ – Tiwa Aina Feb 8 '18 at 15:47
• @Henry yes, this is one of various assumptions which need to be made in order to use the Poisson distribution, and such an assumption is always made whenever the Poisson distribution is used. Luckily for large values of $X$ the probabilities become vanishingly small, which makes such assumptions reasonable. – David Quinn Feb 8 '18 at 15:55
• @DavidQuinn So the Poisson assumptions assume an infinite $N$? How do we reconcile that with the binomial definition if we take the limit as $N \rightarrow \infty$? – Tiwa Aina Feb 8 '18 at 16:35