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A fair coin is tossed until two heads have appeared.

  1. Given that exactly $k$ tosses were required, what is the conditional probability that the first toss resulted in heads?
  2. 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".

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  • $\begingroup$ Well, (first toss heads and k tosses were required)=(first toss heads and k-2 next tosses tails and next toss heads), so what is the problem? $\endgroup$ – Did Sep 22 '16 at 6:52
  • $\begingroup$ @Did Then I am getting (1*(1/2)^(k-2)*1)/(1/2)^k = 4 ...I am sure I am wrong..I don't know where $\endgroup$ – roang Sep 22 '16 at 7:00
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    $\begingroup$ No idea what you are computing in your comment and how you arrive at 4, but anyway, P(first toss heads and k-2 next tosses tails and next toss heads)=1/2^k, obviously. $\endgroup$ – Did Sep 22 '16 at 7:02
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  1. 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?

  1. 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.

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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 $\frac{1}{k-1}$

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  • $\begingroup$ I understood your solution. But the part I could not come up with was {k-1}Choose{1} How do I know that I have to apply this binomial part instead of just calculating the probabilities? $\endgroup$ – roang Sep 23 '16 at 1:08
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    $\begingroup$ @rowang if k tosses are required to get two heads, then that means the kth toss was a head, what we know about the previous k-1 tosses is that exactly one of them was a head. That can occur in exactly k-1 different ways, which is $\binom{k-1}{1}$ - so that's how I calculated the probability of B - the second head appeared on toss k - you might want to consider what the probability of the third head appearing on toss k is, in which case you'd need two heads in k-1 tosses, and $\binom{k-1}{2}$ $\endgroup$ – Cato Sep 23 '16 at 9:01
  • $\begingroup$ If I really understood your explanation this must be correct: From @GrahamKemp answer above to part B pk = P(H2≥k) = 1−P(H2<k) =1-[P(no heads in k-1 tosses) + P(exactly 1 head in k-1 tosses)] = 1- [$1/2^(k-1)$ + $\binom{k-1}{1}$ $1/2^(k-1)$] Correct me if I am wrong. $\endgroup$ – roang Sep 23 '16 at 9:14

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