# A problem on probability involving binomial distribution

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:$$\sum_{i=7}^n Pr(X=i)$$ $$=\sum_{i=7}^n{15\choose i}0.5^i(1-0.5)^{15-i}$$
• $Pr(X=7)={15\choose 7}0.5^7(1-0.5)^{15-7}$ just gives the probability for the amount injured = 7, but not the range of 7 to n, correct? – joe_04_04 Feb 9 '18 at 22:14
• $1-\sum_{i=1}^6 {15\choose i}0.5^i(1-0.5)^{15-i}$? – joe_04_04 Feb 9 '18 at 22:19
• $1-\sum_{i=1}^6 {15\choose i}0.5^i(1-0.5)^{15-i}$ and $=\sum_{i=7}^n{15\choose i}0.5^i(1-0.5)^{15-i}$ are the same thing – Dylan Zammit Feb 9 '18 at 22:21