[Math] Find dependent event when two dice are thrown simultaneously.

diceindependenceprobabilityprobability theory

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 ?

  1. $E_1$
    and $E_3$
    are independent.
  2. $E_1$
    , $E_2$
    and $E_3$
    are independent
  3. $E_1$
    and $E_2$
    are independent.
  4. $E_2$
    and $E_3$
    are independent.

My attempt :

$E_1=(4,1) (4,2) (4,3) (4,4) (4,5) (4,6),$

$E_2 = (1,2) (2,2) (3,2) (4,2) (5,2) (6,2)$

$E_3 = (1,2)……. (6,5)$ One even and one odd

$P(E_1) =1/6$

$P(E_2) = 1/6$

$P(E_3) =1/2$

$P(E_1.E_2.E_3) = 0$

$P(E_1.E_2) = 1/36 //(4,2)$

$P(E_1).P(E_2) =1/6.1/6 =1/36$

$E_1 ,E_2$ are dependent.

But somewhere, answer is given option $(2)$.

Can you explain in formal way, please?

Best Answer

Two events $A$ and $B$ are independent if and only if their joint probability equals the product of their probabilities:

$$\mathrm{P}(A \cap B) = \mathrm{P}(A)\mathrm{P}(B)$$

Here we have that

$$P(E_1 \cap E_2) = \frac{1}{36}$$

$$P(E_1) \cdot P(E_2) =\frac{1}{6} \cdot \frac{1}{6} = \frac{1}{36}$$

So they are independent.

On the other hand

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

$$P(E_1 \cap E_2) = \frac{1}{36}, \:\:P(E_2 \cap E_3) = \frac{1}{12}, \:\: P(E_1 \cap E_3) = \frac{1}{12}$$

$$P(E_1 \cap E_2 \cap E_3) = 0 \neq P(E_1) \cdot P(E_2) \cdot P(E_3)$$

So option $(2)$ is the right answer.

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