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?

My approach:

Is this correct or I am doing something wrong?

• Probability of having the weighted die. Sorry, I edited the question. – meta_finance Jul 8 '16 at 21:08
• The denominator is a bit smudgy. If the second term is $(1/2)(10/36)$ it is right. – André Nicolas Jul 8 '16 at 21:17
• It looks good (though your definition of $B$ isn't quite right). – David Mitra Jul 8 '16 at 21:18
• @AndréNicolas: Yes it is $(1/2)(10/36)$. Thank you. – meta_finance Jul 8 '16 at 21:19
• @DavidMitra: Should $B$ be getting six in one of the throws? – meta_finance Jul 8 '16 at 21:20

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 $$\frac{9}{14}$$