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What is the probability of needing more than $5$ tosses to get $2$ heads for a fair coin?

I know that $E[X] = \frac{1}{p}$ for geometric distribution so the expected number of tosses to get $2$ heads must be $4$. but whats the probability of needing more than $5$ tosses to get $2$ heads ?

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  • $\begingroup$ With regards to your other (now deleted) question, please note that: 1. I did not down-vote it. I merely wrote a comment asking for clarification, and another comment implying that you have misinterpreted of one of the details in your question. The down-vote was most likely due to the fact that you have not demonstrated any attempt to solve the problem on your own. 2. I wrote an answer, but you deleted the question. $\endgroup$ – barak manos Nov 17 '16 at 8:07
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You need more than 5 tosses if you fail to succeed within 5 tosses, i.e., we are looking for the probability of obtaining at most one head with 5 tosses. That's $\frac {1+5}{32}$-

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  • $\begingroup$ Thanks. I did it by subtracting negative binomial from 1. Answer comes out to be the same. $\endgroup$ – Awais Jafar Nov 17 '16 at 6:44
  • $\begingroup$ on a related topic. what would be the conditional probability of getting 2nd head on 5th flip given the second head was not on second flip ? Would it be the same as getting second head on 4th flip ? as geometric is memorless $\endgroup$ – Awais Jafar Nov 17 '16 at 6:46

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