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EEJ MAIN Mathematics QUESTION #6978
Question 1

Two fair six-faced dice $A$ and $B$ are rolled together. Define events:
$E_1$: die $A$ shows 4, $\quad E_2$: die $B$ shows 2, $\quad E_3$: sum of both dice is odd.
Which of the following statements is NOT true?

  • $E_2$ and $E_3$ are independent
  • $E_1$ and $E_3$ are independent
  • $E_1$, $E_2$ and $E_3$ are independent✔️
  • $E_1$ and $E_2$ are independent
Correct Answer Explanation

Total outcomes $= 36$.

$P(E_1)=\frac{6}{36}=\frac{1}{6}$, $P(E_2)=\frac{1}{6}$, $P(E_3)=\frac{18}{36}=\frac{1}{2}$

$E_1$ and $E_2$: $P(E_1\cap E_2)=\frac{1}{36}=\frac{1}{6}\cdot\frac{1}{6}$ ✓ Independent.

$E_1$ and $E_3$: For sum odd with die A showing 4 (even), die B must be odd: 3 outcomes. $P(E_1\cap E_3)=\frac{3}{36}=\frac{1}{12}=\frac{1}{6}\cdot\frac{1}{2}$ ✓ Independent.

$E_2$ and $E_3$: For sum odd with die B showing 2 (even), die A must be odd: 3 outcomes. $P(E_2\cap E_3)=\frac{3}{36}=\frac{1}{12}=\frac{1}{6}\cdot\frac{1}{2}$ ✓ Independent.

$E_1\cap E_2\cap E_3$: Die A=4, Die B=2, sum=6 (even) — impossible! $P(E_1\cap E_2\cap E_3)=0$.

But $P(E_1)\cdot P(E_2)\cdot P(E_3)=\frac{1}{6}\cdot\frac{1}{6}\cdot\frac{1}{2}=\frac{1}{72}\neq 0$.

So $E_1,E_2,E_3$ are not mutually independent (as a triple), even though pairwise independent.