Conditional probability that the first toss resulted in heads 
A fair coin is tossed until two heads have appeared.
  
  
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*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".
 A: 
  
*
  
*Given that exactly $k$ tosses were required, what is the conditional probability that the first toss resulted in heads?
  

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?


  
*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 $p_k \leqslant 0.1$.
  

$$p_k~=~\mathsf P(H_2\geq k) ~=~ 1-\mathsf P(H_2<k)$$
That is: $p_k$ is the probability that you get no more than one head among the first $k-1$ tosses.
A: A = First toss Was a head
B = k tosses required
$P(A|B) = P(A \cap B)/P(B)$
for $P(A \cap B)$ (probability that first toss was a head then k tosses were reuired to get a second head) we need to consider that in this case, toss 1 was a head, and toss k was a head, all tosses in between were tails - therefore it is a a single unique chain of possible tosses, $P = 1 / 2^k$
For P(B), consider that toss k was a head, and tosses 1 to k-1 contained 1 head (at any time) 
$P(B) = \binom{k-1}{1}(1/2^k) =  (k -1) / 2^k$
therefore 
$P(A|B) = P(A \cap B)/P(B)$
$=\frac{( 1 / 2^k)}{((k -1) / 2^k)} = 1/ (k-1)$
answer $\frac{1}{k-1}$
