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A fair coin is tossed. If it comes up heads, dice 1 is rolled three times. If it comes up tails, dice 2 is rolled three times. Dice 1 contains $2$ blue sides and $4$ green sides. Dice 2 contains $3$ blue sides and $3$ green sides. If it is not known which die was used, and the first and second throw showed green, find the probability the third throw will be green.

Attempt:

Using Bayes' rule:

$P(A = \mbox{green on third}|B =\mbox{green on first and second})$,

\begin{equation} P(A|B) = \frac{P(B|A)P(A)}{P(B)}, \end{equation} where $P(B|A)$, $P(A)$, $P(B)$ need to be calculated for both dice. \begin{equation} P(B|A) = \left(0.5\times \frac{4}{6}\times \frac{4}{6}\right) + \left(0.5\times \frac{3}{6}\times \frac{3}{6} \right) \end{equation} \begin{equation} P(A) = (0.5\times \frac{4}{6}) + (0.5\times \frac{3}{6}) \end{equation} \begin{equation} P(B) = \left(0.5\times \frac{4}{6}\times \frac{4}{6}\right) + \left( 0.5\times \frac{3}{6}\times \frac{3}{6}\right) \end{equation}

By this thinking, $P(B)$ = $P(B|A)$. Is this correct?

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    $\begingroup$ Normally Bayes rule is used if $P(A\mid B)$ must be found and it is more easy to find $P(B\mid A)$. But is that the case here? I don't think so. $\endgroup$ – drhab May 30 '18 at 14:48
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Your $P(A)$ and $P(B)$ are correct, but your $P(B|A)$ is not (if it were, then independence would be given, and the opposite is the case here).

Just use the definition $P(B|A)=\frac{P(A\cap B)}{P(A)}$ and use total probability on the numerator to obtain $$P(A\cap B)=\left(0.5\times \frac{4}{6}\times \frac{4}{6}\times\frac{4}{6}\right) + \left(0.5\times \frac{3}{6}\times \frac{3}{6}\times \frac{3}{6} \right)$$ and so $$P(B|A)=\frac{13}{36}.$$


Actually, Bayes is not needed here. Just calculating $P(A|B)=\frac{P(A\cap B)}{P(B)}=\frac{91}{150}$ should be enough.

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  • $\begingroup$ Great - thanks for clearing that up! $\endgroup$ – Sjoseph May 30 '18 at 14:38
  • $\begingroup$ Is it still classed as a conditional probability, if Baye's is not needed? $\endgroup$ – Sjoseph May 30 '18 at 14:50
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    $\begingroup$ @Sjoseph Yes, it's a conditional probability, as is any probability of the form $P(\cdot |\cdot )$. $\endgroup$ – The Phenotype May 30 '18 at 14:57
  • $\begingroup$ If the question was the same except we instead wanted to know was the probability of it being die 1, would it then be a baye's problem? $\endgroup$ – Sjoseph May 30 '18 at 15:08
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    $\begingroup$ @Sjoseph Bayes theorem is used whenever you want to calculate $P(A|B)$, where $P(A|B)$ and $P(A\cap B)$ are "too difficult" to calculate directly and $P(B|A)$ is "easy", so you use the equation $P(A|B) = \frac{P(B|A)P(A)}{P(B)}$. This is exactly what happens in the scenario you are proposing. $\endgroup$ – The Phenotype May 30 '18 at 15:14
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I would not use Bayes rule here.

To be found is $$P(A\mid B)=\frac{P(A\cap B)}{P(B)}$$

Now find $P(A\cap B)$ on base of $$P(A\cap B)=P(A\cap B\mid H)P(H)+P(A\cap B\mid T)P(T)$$

and also find $P(B)$ on base of $$P(B)=P(B\mid H)P(H)+P(B\mid T)P(T)$$

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