Roll 2 dice. Let A = ‘the first die is odd’, B = ‘the second die is odd’, and C = ‘the sum is odd’

Question

I am currently working on a probability problem. I am new to probability (just now studying it). I came across this question just working on some problems.

The question asks, are A and B independent? By intuition I would say so.

A = {1,3,5}, B = {1,3,5}

(Alternatively, I think I could also represent them as pairs with size sets of 36; the probabilities would still be 1/2).

$P(A) = 1/2$

$P(B) = 1/2 $

However, whenever I find the $P(A \cap B)$, it does not equal $P(A)P(B)$.

I get that $$P(A \cap B)=1/2$$ $$|A \cap B|=3,|\Omega|=6$$ where $\Omega$ is the total number of outcomes.

Thus, $$P(A \cap B)=|A \cap B|/|\Omega|=3/6=1/2$$ However, $$P(A)P(B)=1/4$$ Therefore, $$P(A \cap B)=1/2 \neq P(A)P(B)=1/4$$.

This leads to a rejection of independence.

Have I made a mistake?

Answer 1

They are independent. There are $3$ ways for each die to be odd. Therefore there are $3^2 = 9$ ways for both dice in a pair to be odd. There are 6 results for each die. Therefore there are $6^2 = 36$ possible results for a pair of dice. The probability for a pair of dice to both be odd is then $$ P(A \cap B) = \frac{9}{36} = \frac{1}{4} = \frac{1}{2} \cdot \frac{1}{2} = P(A) P(B) $$

The mistake you made is when you wrote $|A \cap B| = 3$. It is actually 9 (and the full space has size 36).

Answer 2

You wrote that $|A \cap B| = 3, |\Omega| = 6$. This is not correct.

$A = \{1,3,5\}, B = \{1,3,5\}$

For throw of two dice, $|\Omega| = 36$

$A \cap B$ is the event of both dice turning up odd. That is,

$A \cap B = \{1,1\}, \{1,3\}, \{1,5\}, \{3,1\}, \{3,3\}, \{3,5\}, \{5,1\}, \{5,3\}, \{5,5\}$

$|A \cap B| = 9$

$\displaystyle |\frac{A \cap B|}{|\Omega|} = \frac{9}{36} = \frac{1}{4}$