Two dice are thrown simultaneously. What is the probability of getting a multiple of $2$ on one die and a multiple of $3$ on the other?

According to me, the answer should be $\frac16$, as the probability of getting a multiple of $2$ is $\frac12$, and the probability of getting a multiple of $3$ is $\frac13$. So the combined probability is $\left(\frac12\right)\left(\frac13\right) = \frac16$.

But the answer in the book says that the probability is $\frac{11}{36}$.

What is the error in my method? How does one solve this problem properly?


Denote event "multiple of $2$" as $A$, event "multiple of $3$" as $B$. We use subscript $1$ and $2$ to denote events for the two dice. For example, $A_1$ denotes event that the first die is multiple of $2$.

The answer to the question is \begin{align} &\Pr((A_1 \cap B_2) \cup (A_2 \cap B_1)) \\ = &\Pr(A_1 \cap B_2) + \Pr(A_2 \cap B_1) - \Pr(A_1 \cap B_2 \cap A_2 \cap B_1)\\ = &\Pr(A_1)\cdot\Pr(B_2) + \Pr(A_2)\cdot\Pr(B_1) - \Pr(A_1\cap B_1)\cdot\Pr(A_2 \cap B_2)\\ = &\frac{1}{2}\cdot\frac{1}{3} + \frac{1}{2}\cdot\frac{1}{3} - \frac{1}{6}\cdot\frac{1}{6}\\ = &\frac{11}{36} \end{align}

Here, the identity $\Pr(A\cup B) = \Pr(A) + \Pr(B) - \Pr(A\cap B)$ is used.

I guess that you only consider the case where the first die is multiple of $2$ and the second die is multiple of $3$. This is not correct.


The error you've made is that you haven't considered the possibility that either dice might have the multiple of 2, or the multiple of 3.

For a problem with these kinds of small numbers we can use fairly weak combinatorial methods.

The total number of outcomes is 36, we count the pairs of numbers such that one is a multiple of 2, and one is a multiple of 3.

The multiples of 2 are 2, 4, 6 the multiples of 3 are 3, 6. So the pairs satisfying the question are

$(2,3)(4,3)(6,3)(3,2)(3,4) (3,6)(2,6)(6,2)(4,6)(6,4) (6,6)$

This is eleven pairs of 36 so $$\frac{11}{36}$$

We could also consider that each multiple of 2 gives 4 pairs (2,3)(3,2)(2,6)(6,2) with three multiples of 2 we have 12 pairs, but we have to discount (6,6) as it is not distinct when we swap the numbers. So we have $$\frac{3 \times 4 -1}{36} = \frac{11}{36}$$

  • $\begingroup$ Please edit: $32 \neq 36$. $\endgroup$ – Dilip Sarwate Jun 4 '15 at 2:38
  • $\begingroup$ edited. thanks for noticing. $\endgroup$ – Howerut Jun 4 '15 at 3:02

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