A fair coin is tossed until two heads have appeared.
- Given that exactly $k$ tosses were required, what is the conditional probability that the first toss resulted in heads?
- If $p_k$ is the probability that at least $k$ tosses are required, find a formula for $p_k$ and find the smallest $k$ such that $p_k\le0.1$.
How do I approach problems like this? For the first question I am not able to apply the Bayes/Price theorem because I am not sure how to derive the $P(A\cap B)$ expression in the numerator. For the second, I am stuck at "at least $k$ tosses".
Best Answer
Yes, you use Bayes' formula: Let $H_n$ count the trials until head $\#n$ occurs.
$$\mathsf P(H_1=1\mid H_2=k) ~=~ \dfrac{\mathsf P(H_2=k\mid H_1=1)~\mathsf P(H_1=1)}{\mathsf P(H_2=2)}$$
Can you now find these probabilities from first principles?
$$p_k~=~\mathsf P(H_2\geq k) ~=~ 1-\mathsf P(H_2<k)$$
That is: $p_k$ is the probability that you get no more than one head among the first $k-1$ tosses.