I'm trying to figure out the method required to solve this problem, so I stripped out the actual values to keep from getting a direct answer.
A random sample of $n$ people who walk to work are chosen, what is the probability that at least $r$ of them are injured, given that the probability of being injured while walking to work is $p$.
I don't know where to go. It feels like a binomial probability problem, but it covers a range of trials and not just one value exactly. My guess was to calculate 1-BinomCDF(n, p, r - 1). Does this seem accurate? For example, if $n = 15$, $r = 7$, $p = 0.5$.
I would have 1-BinomCDF(15, 0.5, 6)
or $$1-\sum_{i=1}^6 {15\choose i}0.5^i(1-0.5)^{15-i}.$$
You are right, this does use the binomial distribution.$$Pr(X=r)={n\choose r}p^r(1-p)^{n-r}$$ That simply gives the probability that "r" events are true from a total of "n" possible events, with the probability of the event happening being "p"
So using your values:$$\sum_{i=7}^n Pr(X=i)$$ $$=\sum_{i=7}^n{15\choose i}0.5^i(1-0.5)^{15-i}$$