Accidents recur in a factory at the rate of 3 per week. Assume that accidents happen randomly and independently of each other.
I've figured out:
What is the probability of no accidents on a given day? Assume the factory works 7 days per week.
$\frac{\left(\frac{3}{7}\right)^0 e^{-\frac{3}{7}}}{0!}$
However, I'm not sure how to figure out the following:
What is the probability of there being 3 accident-free days in a week?
I'm trying something like: $\left(\frac{1}{e^3}\right)^3 \left(1-\frac{1}{e^3}\right) \binom{7}{4}$
But I'm pretty sure im doing it incorrectly.
Use $Y \sim Binomial(7, e^{-\frac{3}{7}})$