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 ?
- $E_1$
and $E_3$
are independent. - $E_1$
, $E_2$
and $E_3$
are independent - $E_1$
and $E_2$
are independent. - $E_2$
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.