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I believe that the probability of the sum is seven is (1/18), unless i'm wrong of course. I'm getting stuck getting the top value. Any help? Is the probability of one of the dice being four = 1/3? Then it would be (1/3)(1/18)/ (1/18)?

2 Answers 2

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Let A be the event that the sum is seven and B be the event that one of the dice is four then.

$P(A \cap B) = \frac{1}{3} \cdot \frac{1}{6}$

where 1/3 comes is the probability of getting either a 3 or a 4 with one of the dice and 1/6 is the probability of getting the appropriate number to add to seven.

$P(A) = 1 \cdot \frac{1}{6}$

Since you can get a seven by having any number on one of the dice with probability 1, and the required number on the other dice with probability 1/6.

Thus,

$P((A \cap B)|A) = \frac{P(A | A \cap B).P(A \cap B)}{P(A)}=\frac{1 \cdot 1/18}{1/6} = \frac{1}{3}$

Using Bayes rule.

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Given that the sum is $7,$ the possibilities are:

$(1,6), (2,5), (3,4), (4,3), (5,2),(6,1)$.

Out of theses, only $(3,4)$ and $(4,3)$ satisfy the condition.

Hence the desired answer is $\frac13$.

  • 3
    For this kind of simple case, enumerating the possibilities is often the fastest solution. – 2017-01-29
  • 0
    This works well since we can think about it as shrinking the sample space to the set of events whose sum equals 7 & then picking two events that contain 4 i.e. (4,3) and (3,4) – 2017-09-22