The probability of a man hitting a target is $2/3$. If he doesn't stop shooting until he hits the target for the first time, a) What is the probability of taking 5 shots to hit the target? b) Which is the least number of shots needed in order to have a probability (of hitting the target) greater than $0.95$?
a) Clearly, a geometric distribution $$P(X = 5) = f(x) = p(1 - p)^{x-1}$$ $$ \frac{2}{3}\left(\frac{1}{3}\right)^4 = 0.00243$$
b) Same as before, now having to solve for $x$ $$P(X=x) = 0.95 = \frac{2}{3}\left(\frac{1}{3}\right)^{x-1}$$ after some algebra, I end up with $$0.475 = \left(\frac{1}{3}\right)^x$$ then I try this: $$log_{1/3}0.475 = x$$ but that doesn't seem to be valid mathematics. Help.
We want the probability of not hitting to be $\lt 0.05$. The probability of not hitting in $x$ trials is $\left(\frac{1}{3}\right)^x$. We want this to be less than $0.05$.
One does not really need theory to find the answer, just a bit of fooling around with the first few powers of $3$. But if we want to use logarithms, we have $\left(\frac{1}{3}\right)^x\lt 0.05$ if and only if $x\ln(1/3)\lt \ln(0.05)$. Now we can find by calculator the solution of $x\ln(1/3)=\ln(0.05)$ and round suitably. (Logarithms to any base will do. You may prefer logarithms to the base $10$.)