I'm having some trouble with this problem
Suppose you flip a biased coin until a head appears. The coin has a $75%$ chance of coming up tails. Let $n$ be the number of flips that you need to do. What is the probability of the following events:
a) $n$ is at most $3$?
b) $n$ is even (note that a geometric series of the form $a + ar + ar^2 + ar^3 + …$ is equal to $\frac{a}{1-r}$
We really haven't talked much about biased coins, or even how to use variables like $n$ in answers.
Best Answer
The biased coin in your case only means that $P(T)=\frac{3}{4}$.
First note that to stop at a single Head in $n$ tosses, you need to get Tail in the first $n-1$ tosses and a Head in the last (that is $n^{th}$ toss).
$1.$ When $n\leq3$. You can get a head in these ways - $H,TH,TTH$.
(Here $TH$ implies first toss gives Tail and second toss gives Head)
So the probability is $$P(H) + P(TH) + P(THH)=\frac{1}{4}+\frac{3}{4}.\frac{1}{4}+\frac{3}{4}.\frac{3}{4}.\frac{1}{4}$$
$2.$ When $n$ is even.
Here you can see that your probablity will be given by $$P(TH)+P(TTTH)+P(TTTTTH)+\dots$$ Write the probabilities in a similar way to the first case and you will obtain an infinite G.P, the formula for sum to which is given in the question.