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I have been going through and doing some (non-assessed) homework questions, but am getting really stuck on conditional probability. The following problem is one that I simply cannot get my head around.

Question: Die A has four red and two blue faces, and die B has two red and four blue faces. One of the dice is selected at random for use.

i). What is the probability of red being thrown?

ii). If the first two throws resulted in red, what is the probability of red for the third throw?

I was able to get i), no trouble, it came out as 1/2. The second part has me entirely lost though.

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  • $\begingroup$ The probability of ii) is 1/2 because they are independent random variables, i.e. every throw is independent of each other. $\endgroup$ – Masacroso Aug 11 '16 at 10:29
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    $\begingroup$ you can use Bayes theorem, you need to find the probability that the red die is in use, so it is P(red die in use given two reds) = P(we select red die and get two reds) / P(2 reds) $\endgroup$ – Cato Aug 11 '16 at 10:30
  • $\begingroup$ @ Masacroso the third throw is using the same die as in 1 and 2, and therefore the 2 throws give us information, in an extreme case the die would give one color only, making it a 100% chance the red die is in place, and a 100% chance of red again $\endgroup$ – Cato Aug 11 '16 at 10:44
  • $\begingroup$ I think P(red dice in use) = (1/2 x 2/3 x 2/3) / (1/4) = (8 /9), so P(red) = (8/9) x (2/3) + (1/9) x (1/3) = 17 / 27 $\endgroup$ – Cato Aug 11 '16 at 10:45
  • $\begingroup$ I understand @AndrewDeighton but the estimation is very weak. In any case we need to apply a hypothesis test to know it significance. $\endgroup$ – Masacroso Aug 11 '16 at 10:50
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For part two, start by considering the question: "What is the probability that two throws result in two reds?"

Since Die A has a $2/3$ probability of providing red, and Die B has a $1/3$ probability of providing red. So:

$P(\text{2 reds}\ |\ A) = (2/3) \times (2/3) = 4/9$

$P(\text{2 reds}\ |\ B) = (1/3) \times (1/3) = 1/9$

The probability for receiving two reds in the first place is half the probability of scoring two reds with die A plus half the probability of scoring two reds with die B, since both dice get selected with equal probability:

$P(\text{2 reds}) = (1/2) \times (4/9) + (1/2) \times (1/9) = (4/18) + (1/18) = 5/18$

Now that we know $P(\text{2 reds})$ and $P(\text{2 reds}\ |\ A)$, we can use Bayes' theorem to compute $P(A\ |\ \text{2 reds})$:

$P(A\ |\ \text{2 reds}) = \frac{P(\text{2 reds}\ |\ A) \times P(A)}{P(\text{2 reds})} = \frac{(4/9) \times (1/2)}{5/18} = \frac{4/18}{5/18} = 4/5$

So we're using die A with $4/5$ probability, die B with $1/5$ probability. Finally, the probability for getting the third red can be calculated:

$P(\text{red}\ |\ \text{2 reds}) = P(A\ |\ \text{2 reds}) \times (2/3) + P(B\ |\ \text{2 reds}) \times (1/3)\\ =(4/5) \times (2/3) + (1/5) \times (1/3) = (8/15) + (1/15) = 9/15 = 3/5$

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Hints:

Let $R_{i}$ denote the event that throw $i$ results in red. Let $A$ denote the event that $A$ is selected for use. Let $B$ denote the event that $B$ is selected for use. You are looking for $\Pr\left(R_{3}\mid R_{1}\cap R_{2}\right)$

  • $\Pr\left(R_{3}\mid R_{1}\cap R_{2}\right)=\frac{\Pr\left(R_{1}\cap R_{2}\cap R_{3}\right)}{\Pr\left(R_{1}\cap R_{2}\right)}$

  • $\Pr\left(R_{1}\cap R_{2}\right)=\Pr\left(R_{1}\cap R_{2}\mid A\right)\Pr\left(A\right)+\Pr\left(R_{1}\cap R_{2}\mid B\right)\Pr\left(B\right)$

  • $\Pr\left(R_{1}\cap R_{2}\cap R_{3}\right)=\Pr\left(R_{1}\cap R_{2}\cap R_{3}\mid A\right)\Pr\left(A\right)+\Pr\left(R_{1}\cap R_{2}\cap R_{3}\mid B\right)\Pr\left(B\right)$

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There is a hidden assumption in the statement "one of the dice is selected randomly before use": the same dice is used in all three throws. If the dice is chosen individually before each throw, as some might interpret it, the throws would be independent and the answer to (ii) $\frac12$, as I had it in an earlier version of the answer.

Here is a derivation of the correct answer that is a little more intuitive. Two reds have shown up. The probability that dice A caused it is $\left(\frac23\right)^2=\frac49$, while the probability that dice B caused it is $\left(\frac13\right)^2=\frac19$. In other words, dice A is 4 times more likely to have been selected than dice B.

The probability that red comes up on the third throw is then the sum of $\frac45\times\frac23$ (for dice A) and $\frac15\times\frac13$ (for dice B), which evaluates to $\frac35$.

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  • $\begingroup$ I disagree, if you had a die with 1 million red faces and 1 blue faces and the other with 1 million blue faces and on red - then if I chose one at random and told you it came up red, it's pretty obvious it's going to be the die with a million red faces, with a massive degree of cetainty, then my next roll is red again $\endgroup$ – Cato Aug 11 '16 at 10:32
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    $\begingroup$ @AndrewDeighton I have rewritten my answer again to explain the source of confusion with this question. $\endgroup$ – Parcly Taxel Aug 11 '16 at 11:50

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