Assume that a gambler has two dice, one of which is fair, and the other is biased toward landing on $six$, so that $0.25$ of the time it lands on $six$, and $0.15$ of the time it lands on each of $1$, $2$, $3$, $4$ and $5$.
Question: The gambler chooses a dice at random and rolls it once and the dice comes up with a 6. What is the probability that fair die was used?
MY WORKING
Let $F$ denote the event that the fair die is chosen.
Let $F'$ denote the event that the biased die is chosen.
Let $X$ denote the event that die comes up with 6
Hence, We need to calcualte $P(F|X)=?$
Applying Bayes' theorem we have:
$P(F|X)=\frac{P(X|F)P(F)}{P(X|F)P(F)+P(X|F')P(F')}$
Now, All we have to do is find the values of each term in the above equation and plug in to get the final answer.
So we have,
$P(F)=\frac{1}{2}$, $P(F')=\frac{1}{2}$
By conditional probability we have:
$P(X|F)=\frac{P(X\cap F)}{P(F)}$ —-> $(i)$
$P(X|F')=\frac{P(X\cap F')}{P(F')}$ —-> $(ii)$
Now, this is the part where I am stuck. I am confused on how to calculate both numerators in $(i)$ and $(ii)$, which are:
$P(X\cap F)$, $P(X\cap F')$. I don't seem to find any relation between both $X$ & $F$ OR $X$ & $F'$ which may reduce the intersection, like one may be subset of the other.
Any guidance will be appreciated. Thanks
Best Answer
You have $P(F|X)=\frac{P(X|F)P(F)}{P(X|F)P(F)+P(X|F')P(F')}$
where $P(F)$ and $P(F')$ can be cancelled since they're equal.
It is thus sufficient to just compare both outcome probabilities, e.g. $\frac{1}{6}$ and $\frac{1}{4}$, over their sum.
$P(F|X)=\frac{P(X|F)}{P(X|F)+P(X|F')}$
$= \frac{\frac{1}{6}}{\frac{10}{24}} = \frac{24}{60} = \frac{2}{5}$
$P(F'|X)=\frac{P(X|F')}{P(X|F)+P(X|F')}$
$= \frac{\frac{1}{4}}{\frac{10}{24}} = \frac{24}{40} = \frac{3}{5}$