A die is thrown two times and the sum of the scores appearing on the die is observed to be a multiple of 4. Then the conditional probability that the score 4 has appeared at least once is

Let us denote by $$A$$ the event “the sum of the two throws is a multiple of 4” and by $$B$$ the event “the score 4 appears at least once”. We wish to find the conditional probability $$P(B\mid A)$$, that is, the probability of $$B$$ given that $$A$$ has already occurred.

By definition of conditional probability we have

$$P(B\mid A)=\dfrac{P(A\cap B)}{P(A)}.$$

First we list every ordered pair $$(x,y)$$, where $$x$$ is the first throw and $$y$$ is the second throw, that belongs to $$A$$. A sum is a multiple of 4 precisely when it equals $$4,8$$ or $$12$$, so we examine these three cases one by one.

For the sum $$4$$:

$$x+y=4\;\Rightarrow\;(x,y)=(1,3),(2,2),(3,1).$$

Hence there are $$3$$ favourable ordered pairs here.

For the sum $$8$$:

$$x+y=8\;\Rightarrow\;(x,y)=(2,6),(3,5),(4,4),(5,3),(6,2).$$

This time we obtain $$5$$ favourable ordered pairs.

For the sum $$12$$:

$$x+y=12\;\Rightarrow\;(x,y)=(6,6).$$

So we get $$1$$ favourable ordered pair in this case.

Summing up all three cases, the total number of elementary outcomes in event $$A$$ is

$$3+5+1=9.$$

Since each ordered pair of two fair die throws is equally likely out of the complete sample space of $$36$$ pairs, we find

$$P(A)=\dfrac{9}{36}=\dfrac{1}{4}.$$

Now we need $$A\cap B$$, the ordered pairs whose sum is a multiple of $$4$$ and that contain at least one $$4$$. Looking back at our list, a score of $$4$$ appears only in the ordered pair $$(4,4)$$ (it belongs to the sum $$8$$ group). No other pair inside $$A$$ contains a $$4$$. Therefore only one outcome satisfies both $$A$$ and $$B$$:

$$(4,4).$$

Consequently, the number of elementary outcomes in $$A\cap B$$ is $$1$$, so

$$P(A\cap B)=\dfrac{1}{36}.$$

Substituting the values of $$P(A\cap B)$$ and $$P(A)$$ into the conditional probability formula, we obtain

$$P(B\mid A)=\dfrac{\dfrac{1}{36}}{\dfrac{1}{4}}=\dfrac{1}{36}\times\dfrac{4}{1}=\dfrac{4}{36}=\dfrac{1}{9}.$$

Hence, the correct answer is Option D.