Question:
Suppose I have either a fair coin or a bent coin, and I don’t know which. The bent coin has a 60% probability of coming up heads.
I throw the coin ten times and it comes up heads 8 times.
Assume I have the bent coin. What is the probability of observing 8 heads in 10 throws, given the bent coin parameter? (In Bayesian Inverse probability problems, this is also called the “likelihood” function of the parameter).
My solution/logic:
Using binomial theorem. $\binom{10}{8}\cdot0.6^8\cdot0.4^2=0.12$
Am I wrong ?
This is correct, except that we should be clear about the meaning of the word "likelihood": The likelihood is that probability as a function of the unobservable parameter $p$ with the data held fixed (i.e. the number of successes in $10$ trials), thus it is $$ p \quad \mapsto \quad {_{10}}C_8 \,p^8 (1-p)^2, $$ and in this case there are only two posible values of $p,$ which are $0.5$ and $0.6.$ Thus you could actually represent the likelihood as a pair: $$ \left( {_{10}}C_8\, 0.5^8 0.5^2 , {_{10}}C_8 \, 0.6^8 0.4^2 \right) = ( 0.043945\ldots , 0.12093235\ldots ) $$