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There're 3 dice, two are fair, one is not. The non-fair dice has an occurrence of $5$ instead of $4$. That is the non-fair dice is $\{1,2,3,5,5,6\}.$ We choose one dice from the three randomly, and throw it twice. Suppose a dice was chosen such that it returned $5$ in both rolls. We now choose a dice from the remaining dice and throw it once. What is the probability that the dice will return $5$?

Let $A_{55}$ be the event in which both rolls returned $5$. In the related question I posted earlier (outcome from fair and non-fair dice, dice are chosen from other dice part1): $$A_{55} = \frac{2}{3}\cdot\frac{1}{6}\cdot\frac{1}{6} + \frac{1}{3}\cdot\frac{1}{3}\cdot\frac{1}{3}$$ Let $A_5$ be the event in which a roll will return $5$. $$ A_5 = \frac{1}{2}\cdot\frac{1}{6} + \frac{1}{2}\cdot \frac{1}{3} $$ because now we need to choose one dice from 2.

Now I'd use Bayes Rule: $$ P(A_5|A_{55}) = \frac{P(A_5 \cap A_{55})}{P(A_{55})} $$ This is pretty much where I'm stuck because I don't know how to proceed with the nominator. $P(A_5)$ and $P(A_{55})$ are not independent so we can't multiply them I think.

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  • $\begingroup$ I am not understanding A55. Can you explain. $\endgroup$ Dec 26, 2016 at 11:32
  • $\begingroup$ $A_5$ should be $\frac23\cdot\frac16+\frac13\cdot\frac13$, shouldn't it? $\endgroup$ Dec 26, 2016 at 11:33
  • $\begingroup$ Doesn't $A_{55}$ take care of the third dice and in $A_5$ we're choosing one dice out of 2? $\endgroup$
    – Yos
    Dec 26, 2016 at 11:35

2 Answers 2

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You found $P(A_{55})$ correctly. Now, we just need to find $P(A_5 \cap A_{55})$. This is a bit complicated since there are more than two cases, but it is still doable.

Case 1: Fair first, then unfair

  • Probability of picking fair is $\frac 2 3$
  • Probability of picking unfair after picking fair is $\frac 1 2$
  • Probability of fair rolling two fives is $(\frac{1}{6})^2=\frac 1 {36}$
  • Probability of unfair rolling five is $\frac 1 3$

Case 2: Unfair first, then fair

  • Probability of picking unfair is $\frac 1 3$
  • Probability of picking fair after picking unfair is $\frac 2 2=1$
  • Probability of unfair rolling two fives is $(\frac{1}{3})^2=\frac 1 {9}$
  • Probability of fair rolling five is $\frac 1 6$

Case 3: Fair first, then fair

  • Probability of picking fair is $\frac 2 3$
  • Probability of picking fair after picking fair is $\frac 1 2$
  • Probability of fair rolling two fives is $(\frac{1}{6})^2=\frac 1 {36}$
  • Probability of fair rolling five is $\frac 1 6$

Case 4: Unfair first, then unfair

  • Impossible because there is only one unfair die, so completely ignore this case

Now, by Law of Total Probability, multiply within the cases and add between:

$$P(A_5 \cap A_{55})=\frac 2 3\cdot\frac 1 2\cdot\frac 1 {36}\cdot\frac 1 3+\frac 1 3\cdot1\cdot\frac 1 9\cdot\frac 1 6+\frac 2 3\cdot\frac 1 2\cdot\frac 1 {36}\cdot\frac 1 6=\frac{7}{648}$$

Now, use Bayes' Rule: $$P(A_5 | A_{55})=\frac{P(A_5 \cap A_{55})}{P(A_{55})}=\frac{\frac{7}{648}}{\frac{2}{3}\cdot\frac{1}{6}\cdot\frac{1}{6} + \frac{1}{3}\cdot\frac{1}{3}\cdot\frac{1}{3}}=\frac{7}{36}$$

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  • $\begingroup$ I think in case 2 you meant "probability of unfair rolling two fives is $\frac{1}{3}\cdot \frac{1}{3}$" and "probability of fair rolling one five is $\frac{1}{6}$" $\endgroup$
    – Yos
    Dec 27, 2016 at 14:41
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    $\begingroup$ @Yos Thanks for the catch! I've edited my answer to fix the typos. $\endgroup$ Dec 27, 2016 at 14:50
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In our class we learnt another approach to dealing with such problems which in my opinion is much simpler. The approach is to draw trees of all possible cases from each event and writing probabilities for each possibility. At the end we multiply the probabilities along the possible "routes" and then add the probabilities of all "routes". Addition of routes is allowed because of the Law of Total Probability. Multiplication along the routes is the direct result of the definition of conditional probability: $P(A|B) = \frac{P(A \cap B)}{P(B)} \Rightarrow P(A \cap B) = P(A|B) \cdot P(B)$

Thus we just need to make sure to have drawn all possible developments of events and then follow the tree for multiplications and additions. Below is the tree which shows possible developments of $P(A_5 \cap A_{55})$ and its result will be the same as the one posted by Noble Mushtak. In the tree $A$ is the throw of fair dice while $B$ is the throw of non-fair dice. $5$ is the roll of 5 on a given throw. Between the circles which represent events their respective probabilites are given. enter image description here

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