If I have two dice, one regular and one loaded. The loaded die has the probability 1/2 of landing a six and rest of the numbers are equally probable. If you select a die randomly and throw it and it shows 6 in one of the throws and not a six in other. What is the probability of having a weighted die?
Is this correct or I am doing something wrong?
Best Answer
Let's denote the events
LD - picked up the loaded dice;
S - shows 6 in the 1st throw and not 6 on the second throw;
We are asked to find $$ P(LD|S) $$ Use Bayesian theorem $$ P(LD|S) = \frac{P(S|LD)P(LD)}{P(S|LD)P(LD) + P(S|\overline{LD})P(\overline{LD})} $$ where $$P(S|LD) = \frac{1}{2}\frac{1}{2},~P(LD)=P(\overline{LD})=\frac{1}{2},~P(S|\overline{LD}) = \frac{1}{6}\frac{5}{6} $$ plug all back to the fraction, I got $$ \frac{9}{14} $$