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?
2026-03-29 16:01:21.1774800081
Biased coin probability calculation
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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 $$ \frac6{26}=\frac3{13} $$