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I have a biased coin with 2/3 chance of heads (and thus 1/3 chance of tails). Question is: given that there was at least one head in 3 flips, what is the probability that there is only 1 head?

How would I solve this?


marked as duplicate by lulu, quid, E. Joseph, Michael Albanese, Vladimir Vargas Dec 12 '16 at 23:16

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  • $\begingroup$ Divide the probability of there being exactly 1head over the probability of there being at least 1head. Also, the probability of there being at least 1 head is 1- the probability of there being no heads. $\endgroup$ – Bram28 Dec 12 '16 at 19:55

The way to set this up is that there might be $0,1,2$ or $3$ heads in the three flips, but we can discard the case of $TTT$ since that has no heads.

Then the a priori probabilities (ignoring the constraint that there was at least one head) are:

  • $HHH: \left(\frac23\right)^3 \cdot \binom{3}{3} = \frac {8}{27}$

  • $HHT, HTH, THH: \left(\frac23\right)^2 \left(\frac13\right)^1 \cdot \binom{3}{2} = \frac {12}{27}$

  • $HTT, THT, TTH: \left(\frac23\right)^1 \left(\frac13\right)^2 \cdot \binom{3}{1} = \frac {6}{27}$

The total is $\frac{26}{27}$ out of which only $\frac6{27}$ have exactly one head, so the answer is $$ \frac6{26}=\frac3{13} $$

  • $\begingroup$ Shouldn't it be $\frac{6}{26}$? $\endgroup$ – paw88789 Dec 12 '16 at 20:42
  • $\begingroup$ Im getting $$\frac{3}{13}$$ .. same as above answer. $\endgroup$ – K Split X Dec 12 '16 at 21:18
  • $\begingroup$ Yes, as Mark himself said, the TTT should be discarded, so $\frac{6}{26} = \frac{3}{13}$. $\endgroup$ – Bram28 Dec 12 '16 at 21:49
  • $\begingroup$ Yes, you are right @paw88789. I will make the change. $\endgroup$ – Mark Fischler Dec 13 '16 at 22:00

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