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We have 2 dice. One is fair. The other one lands by the following probabilities:

6: 1/2
5: 1/10
4: 1/10
3: 1/10
2: 1/10
1: 1/10

We roll both dice.

What is the probability that the sum is 9 given that exactly one die lands on 6?


Here's what I have so far:

Cases where sum is 9:

Case1: Loaded = 6 | Fair = 3 -> 1/2 * 1/6
Case2: Loaded = 3 | Fair = 6 -> 1/10 * 1/6
Case3: Loaded = 5 | Fair = 4 -> 1/10 * 1/6
Case4: Loaded = 4 | Fair = 5 -> 1/10 * 1/6
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                  P(sum is 9) = 2/15

P(exactly one die lands on 6) = P(loaded lands on 6) + P(fair lands on 6) - P(both land on 6)

P(exactly one die lands on 6) = 1/2 + 1/6 - (1/2 * 1/6) = 7/12


P(sum is 9 AND exactly one die lands on 6)
= (P(case1) + P(case2)) / P(sum is 9)
= ((1/2 * 1/6) + (1/10 * 1/6)) / (2/15)
= 3/4

P(sum is 9 | exactly one die lands on 6)
= P(sum is 9 AND exactly one die lands on 6) / P(exactly one die lands on 6)
= (3/4) / (7/12)
> 1

What am I doing wrong?

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2 Answers 2

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The first mistake is when you calculated the probability that exactly one die lands on $6$, as you included the case when both dice are $6$ twice (first when the loaded die is $6$, and secondly when the fair die is $6$) - so you need to subtract the case when they were both $6$ two times:-

$$\begin{align}P(\text{exactly one die lands on 6}) &= P(\text{loaded lands on 6}) + P(\text{fair lands on 6}) -\color{red}{2} P(\text{both land on 6})\\&=\frac{1}{2}+\frac{1}{6}-\frac{2}{12}=\frac{1}{2}\end{align}$$

The second mistake (as Andrew has rightly pointed out) is that you did an erroneous division to calculate $P(\text{sum is 9 AND exactly one die lands on 6})$.

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  • $\begingroup$ Thank you! I ran a simulation to verify that the answer is correct after fixing both errors. $\endgroup$
    – bkoodaa
    Feb 6, 2016 at 20:53
  • $\begingroup$ You are most welcome. $\endgroup$ Feb 6, 2016 at 20:55
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In short, you erroneously divided by $\Pr\{\text{sum is 9}\}$ when calculating $\Pr\{\text{sum is 9 and exactly one die lands on 6}\}$.

$$ \begin{align*} \Pr\{\text{sum is 9 and exactly one die lands on 6}\} & = \Pr\{(6,3),(3,6)\} \\ & = \Pr\{(6,3)\} + \Pr\{(3,6)\} \\ & = \left(\frac{1}{2}\right)\left(\frac{1}{6}\right) + \left(\frac{1}{10}\right)\left(\frac{1}{6}\right) \\ & = \frac{1}{10} \end{align*} $$

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  • $\begingroup$ Thank you. I would upvote if I had enough rep to do it. $\endgroup$
    – bkoodaa
    Feb 6, 2016 at 20:51

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