If a council decides to perform a water quality check every day. And the test costs \$500 each day. If a pollutant is found in the water, they will find the polluter and fine them \$50,000. What does the probability to find a pollutant in the sample each day at least have to be, for the council to be 90% sure that they will not lose money on these tests, in a non-leap year?
I need any hint to solve this question , thanks
This is a difficult question since measurements are not independent (i.e. the results today clearly have an effect on the results tomorrow)
Assuming independence, we have a binomial $Bin(365,p)$ distribution (a yes/no event that repeats $365$ times with a probability of "yes" being $p$)
For the council not losing money, we would need to get at least 4 positive results (the total yearly costs are $182,500$), so we need to find $p$ such that:
$$P(X \geq 4) = 0.9$$
Where $X$ follows the said distribution (if $p$ is higher than that threshold, the council will win money. Otherwise, they will lose it. Equivalently, we need to calculate for which value of p:
$$P(X \leq 3) = 0.1$$
Now (see formulas for Binomial distribution):
$$P(X=0) = (1-p)^{365}$$ $$P(X=1) = 365 \cdot p \cdot (1-p)^{364}$$ $$P(X=2) = \frac{365 \cdot 364}{2} \cdot p^2 \cdot (1-p)^{363}$$ $$P(X=3) = \frac{365 \cdot 364 \cdot 363}{3} \cdot p^3 \cdot (1-p)^{362}$$
Now add those four quantities, equal to $0.1$ and (numerically?) solve for $p$
WARNING: Do not try to compute this with normal approximations, since the value of $p$ could be very small and therefore the approximation way too far off.