Let two fair six-faced dice $$A$$ and $$B$$ be thrown simultaneously. If $$E_1$$ is the event that die $$A$$ shows up four, $$E_2$$ is the event that die $$B$$ shows up two and $$E_3$$ is the event that the sum of numbers on both dice is odd, then which of the following statements is not true?

We begin by noting that each of the two dice is fair and six-faced, so the ordered pairs of outcomes $$\bigl(A,B\bigr)=(1,1),(1,2),\dots,(6,6)$$ are all equally likely. There are in total $$6\times 6=36$$ such pairs, each having probability $$\dfrac1{36}$$.

The three events are defined as follows:

$$E_1:\;A=4,\qquad E_2:\;B=2,\qquad E_3:\;A+B\text{ is odd.}$$

First, we compute the individual probabilities.

For $$E_1$$ we require the first die to show four. The second die may show any of the six numbers, giving $$6$$ favourable pairs: $$(4,1),(4,2),(4,3),(4,4),(4,5),(4,6).$$ Hence $$P(E_1)=\dfrac6{36}=\dfrac16.$$ Exactly the same reasoning applies to $$E_2$$, so $$P(E_2)=\dfrac16.$$ For $$E_3$$, the sum is odd exactly when one die is even and the other odd. There are $$3$$ even numbers and $$3$$ odd numbers on a die, so the number of favourable pairs is $$3\times 3+3\times 3=18.$$ Thus $$P(E_3)=\dfrac{18}{36}=\dfrac12.$$

Now we examine pairwise intersections in order to check pairwise independence.

Intersection of $$E_1$$ and $$E_2$$. Both $$A=4$$ and $$B=2$$ must hold, leaving the single outcome $$(4,2).$$ Therefore $$P(E_1\cap E_2)=\dfrac1{36},\qquad P(E_1)\,P(E_2)=\dfrac16\cdot\dfrac16=\dfrac1{36}.$$ Because these two numbers are equal, $$E_1$$ and $$E_2$$ are independent.

Intersection of $$E_1$$ and $$E_3$$. We already have $$A=4$$ (an even number). For the sum to be odd, $$B$$ must be odd, i.e. $$B=1,3,5.$$ The three favourable outcomes are $$(4,1),(4,3),(4,5).$$ Thus $$P(E_1\cap E_3)=\dfrac3{36}=\dfrac1{12},\qquad P(E_1)\,P(E_3)=\dfrac16\cdot\dfrac12=\dfrac1{12}.$$ The equality confirms that $$E_1$$ and $$E_3$$ are independent.

Intersection of $$E_2$$ and $$E_3$$. Now $$B=2$$ (even), and we need the sum to be odd, so $$A$$ must be odd: $$A=1,3,5.$$ The three favourable pairs are $$(1,2),(3,2),(5,2).$$ Hence $$P(E_2\cap E_3)=\dfrac3{36}=\dfrac1{12},\qquad P(E_2)\,P(E_3)=\dfrac16\cdot\dfrac12=\dfrac1{12}.$$ Again the two values match, so $$E_2$$ and $$E_3$$ are independent.

So far, every pair of events is independent. To see whether all three events are mutually independent, we must consider the triple intersection.

Intersection of $$E_1,$$ $$E_2,$$ and $$E_3$$. Imposing $$A=4$$ and $$B=2$$ fixes the only possible outcome to be $$(4,2).$$ However, for this outcome the sum is $$4+2=6,$$ which is even, not odd. Therefore no single ordered pair satisfies all three conditions simultaneously, and $$P(E_1\cap E_2\cap E_3)=0.$$ On the other hand, $$P(E_1)\,P(E_2)\,P(E_3)=\dfrac16\cdot\dfrac16\cdot\dfrac12=\dfrac1{72}.$$ Because $$0\neq\dfrac1{72},$$ the three events $$E_1,E_2,E_3$$ are not mutually independent.

We have now gathered all the information required to assess the given statements:

• Statement A: “$$E_1$$ and $$E_3$$ are independent.” — True.
• Statement B: “$$E_1$$, $$E_2$$ and $$E_3$$ are independent.” — Not true (they are not mutually independent).
• Statement C: “$$E_1$$ and $$E_2$$ are independent.” — True.
• Statement D: “$$E_2$$ and $$E_3$$ are independent.” — True.

Hence, the correct answer is Option B.