Probability that the sum of the results of two fair dice is $6$, given that one of them shows $4$

Question

I haven't touched probability for a long time, and now I'm a bit rusty.

I am trying to understand this very basic question:

Given two fair dice, what is the probability that the sum of their results is $6$, given that one of the dice shows $4$?

The official answer is $\frac{2}{11}$. How come?

Answer 1

Let $(a,b)$ mean that $a$ came on the first dice and $b$ came on the second dice during a roll. It is given that one of these is a $4$.

The sample space consists of $11$ elements : $(4,1),(4,2),(4,3),(4,4),(4,5),(4,6)$ and everything flipped except for the $(4,4)$, which comes the same when flipped, so we don't count it twice. Hence there are $11$ elements, and out of these, two sum to $6$, namely $(4,2)$ and $(2,4)$. So the answer is $\frac{2}{11}$.

Answer 2

I preassume that at least one of the dice shows a $4$. If it must be exactly then see the answer of Andrew (not $\frac2{11}$ but $\frac2{10}$ because possible outcome $(4,4)$ falls away).

The following equations enable you to find the conditional probability: $$\Pr(D_1+D_2=6\mid D_1=4\vee D_2=4)$$ and that is what you are looking for.

• $\Pr(D_1+D_2=6\mid D_1=4\vee D_2=4)\Pr(D_1=4\vee D_2=4)=\Pr(D_1+D_2=6\wedge(D_1=4\vee D_2=4))$

• $\Pr(D_1=4\vee D_2=4)=1-\Pr(D_1\neq 4\wedge D_2\neq4)=1-\Pr(D_1\neq 4)\Pr(D_2\neq 4)$

• $\Pr(D_1+D_2=6\wedge(D_1=4\vee D_2=4))=\Pr(D_1=4\wedge D_2=2)+\Pr(D_1=2\wedge D_2=4)=\Pr(D_1=4)\Pr(D_2=2)+\Pr(D_1=2)\Pr(D_2=4)$

Denoting the conditional probability by $p$ we find:$$p\left(1-\frac{25}{36}\right)=\frac2{36}$$leading to:$$p=\frac2{11}$$

The answers of Alec and ACTOH are more elegant and direct. See this as invitation to have a look at the theory of conditional probabilities.

Answer 3

There are 11 possible ways of having at least one 4. 11 and not 12 because we don't want to count $(4,4)$ twice.

There are two possible ways of having one 4, and the other number being 2 (which you need for the sum to be 6), namely $(4,2), \ (2,4)$. So two desirable outcomes.

And as always, the probability of the desired outcome, is the number of desirable outcomes, divided by the number of possible outcomes, so that's $\frac2{11}$.

Answer 4

it's P(A|B) or prob of A given B - bayes therorem

P(A|B) = P(A and B) / P(B)

A is total of 6 B is 4 is rolled

hmm, badly worded question - is it one 4 or at least one 4 - anyhow it says 'one' so to me that rules out zero and two

P(A and B) = 2 / 36 = 1 / 18 ({4,2} or {2,4})

P(B) = 2 x (1/6) x (5/6) = 10/ 36 = 5 / 18

P(A|B) = P(A and B) / P(B) = (1 / 18) / (5/18) = 1 / 5

---

allowing 'two' fours then P(B) = 2/6 - 1/36 = 11 / 36

then P(A|B) = P(A and B) / P(B) = (1 / 18) / (11/36) = 2 / 11

Which is there answer. The question says 'one' of the dice shows a four, that excludes two of the dices showing four, just as it excludes zero of the dices showing four - what do other people think?