Let's say you eat at some very cheap restaurant and every time you do there's a 5% chance you'll get food poisoning.
How do you calculate the probability of getting food poisoning if you eat there x days in a row? Obviously eating there 20 days in a row doesn't give you a 100% chance, but I'm really not sure how that kind of probability works.
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$\begingroup$The probability of getting sick the first day is 5%, clearly. The probability of getting sick the first time on the 2nd day would be (.95)(.05). Do you see why? Then you could sum up the probability of the first 20 days this way to see the probability of getting sick any of those days.
Alternately, and more simply, what is the probability of not getting sick for 20 straight days? Then subtract that from 1 to see the probability of not making it 20 straight days.
Another way to look at it is with this diagram. The probability of getting sick on any day is multiplying out the path to get to that day. Each horizontal level represents a new day.
You're looking at a geometric distribution. A "success" would be food poisoning (ironic, huh?):
$$P(X\le 20) = \sum_{k=1}^{20} (1-p)^{k-1}p = 1-(1-p)^{20}$$ Where $p=0.05$ is your probability of success.
This is basically the formulation "What is the probability you'll get food poisoning within $20$ days of eating at the restaurant?".
$\endgroup$ 1 $\begingroup$The probability of not getting food poisoning on a given day is $95\%$ .
The probability of not getting food poisoning on both of two day is $90.25\%$ . (That is $ 0.95 \times 0.95$ .)
The probability of getting food poisoning on at least one of two day is $9.75\%$ . (That is $1-0.9025$ .)
The probability of not getting food poisoning on each of $n$ days is $(0.95)^n$ .
The probability of getting food poisoning on at least one of $n$ days is $1-(0.95)^n$ .
The probability of getting food poisoning on at least one of $20$ days is approximately $87{.\small 84}\%$
It is inadvisable to eat at this restaurant.
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