4
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ 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
0
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.