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?

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

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


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