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Charles has two six-sided dice. One of the dice is fair, and the other die is biased so that it comes up six with probability $\frac{2}{3}$ and each of the other five sides has probability $\frac{1}{15}$ Charles chooses one of the two dice at random and rolls it three times. Given that the first two rolls are both sixes, the probability that the third roll will also be a six is $\frac{p}{q}$ where $p$ and $q$ are relatively prime positive integers. Find $p + q$.

Case 1: Fair die chosen:

$$P = \binom{3}{3} \frac{1}{6^3} \cdot \frac{5^0}{6^0} = \frac{1}{216}$$

Case 2: Unfair die chosen:

$$P_2 = \binom{3}{3} \frac{2^3}{3^3} = \frac{8}{27}$$

$$P + P_2 = \sum P = \frac{1728 + 27}{5832} = \frac{1755}{5832}$$

But that is incorrect, why?

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    $\begingroup$ Hint: Knowing that you got two 6's makes it far more probable that you are working with the loaded die. Indeed that outcome is 16 times more probable with the unfair die. That fact has to be used somewhere in your calculation. $\endgroup$ – lulu Jul 19 '15 at 16:31
  • $\begingroup$ You seem to calculate of both dice the probability that the $3$ times a $6$ will appear. In the question they are talking about another (conditional) probability. Secondly: you are adding up the probabilities, which can only bear any fruit if you are dealing with disjoint events. $\endgroup$ – drhab Jul 19 '15 at 16:37
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The answer you wrote down doesn't capture the information gained from getting two sixes. This problem requires conditional probability. Let die $A$ be the event that the fair die was selected and $B$ be the event that the unfair die was selected. Then we have \begin{align} \mathbb{P}(\text{another } 6 \text{ is rolled }| \text{ two } 6\text{'s were rolled}) &= \mathbb{P}(\text{another } 6 \text{ is rolled }|A)\cdot \mathbb{P}(A| \text{ two } 6\text{'s were rolled}) \\ &+ \mathbb{P}(\text{another } 6 \text{ is rolled }|B)\cdot \mathbb{P}(B| \text{ two } 6\text{'s were rolled}) \\ &= \frac{1}{6}\cdot \mathbb{P}(A| \text{ two } 6\text{'s were rolled}) \\ &+ \frac{2}{3}\cdot \mathbb{P}(B| \text{ two } 6\text{'s were rolled}) .\end{align}

We now use Bayes' theorem to calculate $ \mathbb{P}(A| \text{ two } 6\text{'s were rolled})$ and the similar probability for $B$. We have \begin{align} \mathbb{P}(A| \text{ two } 6\text{'s were rolled})&= \frac{\mathbb{P}(A)\mathbb{P}(\text{ two } 6\text{'s were rolled } | A)}{\mathbb{P}(A)\mathbb{P}(\text{ two } 6\text{'s were rolled } | A) + \mathbb{P}(B)\mathbb{P}(\text{ two } 6\text{'s were rolled } | B)} \\ &= \frac{\frac{1}{36}\cdot\frac{1}{2}}{\frac{1}{36}\cdot\frac{1}{2} + \frac{4}{9}\cdot\frac{1}{36}} \\ &= \frac{1}{17}.\end{align}

Thus, we also have $ \mathbb{P}(B| \text{ two } 6\text{'s were rolled}) =\frac{16}{17}.$ Putting it all together gives $$ \mathbb{P}(\text{another } 6 \text{ is rolled }| \text{ two } 6\text{'s were rolled}) = \frac{1}{6}\cdot \frac{1}{17} + \frac{2}{3}\cdot\frac{16}{17} = \frac{65}{102}.$$

Thus, your answer is $65 + 102 = 167$.

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  • $\begingroup$ Why did you use conditional probability? couldnt you have used casework? $\endgroup$ – Amad27 Jul 19 '15 at 17:15
  • $\begingroup$ You could do (a lot of casework), but the whole idea is that you have to somehow take into account the two rolls that were rolled before. Even if you break it down into a whole bunch of cases, you'd still have to use some kind of conditional statement at some point. $\endgroup$ – Marcus M Jul 19 '15 at 17:46

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