let $X$ r.v. denote the time needed for the event of a cup breaking with exp$(\frac{1}{24})$ distribution, in other words a cup breaks every 24 months(expected value).
If I want to know what is the probability of at most $k$ cups breaking out of $n$ total cups in a year, how would I formulate that?
What I understand is this:
P$(X\leq12)$=$1-(1-e^{-\frac{12}{24}})=e^{-0.5}\approx 0.606$
is the probability that one cup breaks in a year, I am having problem understanding how do I define an inequality for less than equal 3 cups and same time restrict the rate of the exponential function to 12 months.
Is it just:
$1$ cup $=$ $24$ months $\rightarrow $3cups$ =72$ month
so, our new rate is $\lambda = \frac{1}{72}$
then, answer would be: P$(Y<12)$, where $Y$ stands for 3 cups breaking, but then that would only consider 3 cups breaking not at most 3.
Or is it by reducing the rate into : $\frac{1}{12}$ and somehow changing our axis into cups instead of time?
Any help is appreciated as I got really confused here with the exponential dist.
Assuming that the breaking or otherwise of each individual cup occurs independently of any of the others, the number out of $\ n\ $ cups that break in a year will be binomially distributed with parameters $\ n\ $ and $\ p=e^{-0.5}\ $: $$ P(\text{exactly }j \text{ cups break})={n\choose k}e^{-0.5 j}\left(1-e^{-0.5}\right)^{n-j}\ . $$ Therefore $$ P(\text{at most }k \text{ cups break})=\sum_{j=0}^k{n\choose k}e^{-0.5 j}\left(1-e^{-0.5}\right)^{n-j}\ . $$