# A problem on probability involving binomial distribution. 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.

moidu13x8 2022-09-15 Answered
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 $1-BinomCDF\left(n,p,r-1\right)$. Does this seem accurate? For example, if $n=15$, $r=7$, $p=0.5$.
I would have $1-BinomCDF\left(15,0.5,6\right)$ or
$1-\sum _{i=1}^{6}\left(\genfrac{}{}{0}{}{15}{i}\right){0.5}^{i}\left(1-0.5{\right)}^{15-i}.$
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## Answers (1)

ko1la2h1qc
Answered 2022-09-16 Author has 18 answers
Step 1
You are right, this does use the binomial distribution.
$Pr\left(X=r\right)=\left(\genfrac{}{}{0}{}{n}{r}\right){p}^{r}\left(1-p{\right)}^{n-r}$
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"
Step 2
So using your values:
$\sum _{i=7}^{n}Pr\left(X=i\right)$
$=\sum _{i=7}^{n}\left(\genfrac{}{}{0}{}{15}{i}\right){0.5}^{i}\left(1-0.5{\right)}^{15-i}$

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