Suppose I roll a 4-sided die, then flip a fair coin a number of times corresponding to the die roll. Given that i got three heads on the coin flip, what is the probability the die score was a 4?
I recognize that Bayes Formula can be used to solve this but I'm a little stuck on how to apply it. I figured that because the coin flip got three heads, it's impossible the die score is 2 or less, so the die score MUST be greater than 2. But I'm not sure if that really applies.
Answer
I recognize that Bayes Formula can be used to solve this but I'm a little stuck on how to apply it. I figured that because the coin flip got three heads, it's impossible the die score MUST be greater than 2. But I'm not sure if that really applies.
Right, given that you have three heads, you must have rolled 3 or 4 on the die, and flipped the coin so many times. Letting $X$ denote the die result and $Y$ count the heads.
$\mathsf P(X=4\mid Y=3) ~{~=~ \dfrac{\mathsf P(Y=3\mid X=4)\mathsf P(X=4)}{\mathsf P(Y=3\mid X=3)\mathsf P(X=3)+\mathsf P(Y=3\mid X=4)\mathsf P(X=4)}\\~=~ \dfrac{4\cdot 2^{-4}}{2^{-3}+4\cdot 2^{-4}}\\~=~\dfrac 23}$
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