What is the likelihood of a fair coin given 10 heads (with added component)?

Question

What is the likelihood of a fair coin given that it has landed heads up 10 times?

You have a fair coin or a double-headed coin...

$\mathsf P($Fair$\mid 10$ heads$) = \dfrac{(1/2)(1/2^{10})}{(1+2^{10})/(2\cdot 2^{10})}= \dfrac{(1/2)(1/1024)}{1025/2048}$

Is this the correct procedure for setting up this problems?

You have a fair coin or $70\%$ weighted coin favoring heads...

What is the likelihood of a fair coin given that it has landed heads up 10 times?

Answer 1

Lacking other information, we must assume equal prior probability of fairness or $70\%$-bias (towards heads), and that the selection of the coin are mutually exclusive and exhaustive.

The events are: $E$: evidence (10 heads in a row), $F$: Fair coin selected, $F^c$: Biased coin selected.

$\begin{align} \mathsf P(F\mid E) & = \frac{\mathsf P(E\mid F)\mathsf P(F)}{\mathsf P(E\mid F)\mathsf P(F)+\mathsf P(E\mid F^c)\mathsf P(F^c)} \\[1ex] & = \frac{\mathsf P(E\mid F)}{\mathsf P(E\mid F)+\mathsf P(E\mid F^c)} & \because \mathsf P(F)=\mathsf P(F^c) \\[2ex] & \vdots \\[2ex] & = \underline{\qquad} \end{align}$

Substitute and solve.

Answer 2

You want to use bayes theorem where we would have letting $F=$ fair coin and $T$= ten heads came up. thus we have $$P(F|T)=\frac{P(T|F)P(F)}{P(T|F)P(F)+P(T|F^c)P(F^c)}$$ where $P(F)=\frac{1}{2}$ and find $P(T|F)=\left(\frac{1}{2}\right)^{10}$ and $P(T|F^{c})=\left(\frac{7}{10}\right)^{10}$