# Independence of the sum of two dice given one of them is odd the result

I'm not sure how to argue this.

Two fair dice, one red and one blue are rolled. Let $A$ be the event that the number rolled on the red die is odd. Let $B$ be the event that the number rolled on the blue die is odd. Let $C$ be the event that the sum of the two dice is odd. Show that $B$ and $C$ are independent.

It's obvious that $A$ & $B$ are independent I tried $$P(C \mid B) = P(\text{Die Red + Die Blue is odd} \mid B) = \frac{P(\text{Die Red + Die Blue is odd} \cap B)}{P( B )}$$ If $B$ is odd, Die Red should be Even. Therefore, the sum will be Odd. $$\frac{P(A^{c}B)}{P(B)} = \frac{P(A^{c})P(B)}{P(B)}= \frac{1}{2}$$

If someone has a suggestion, I would really appreciate that.

• Isn't that just as simple as applying the definition? $P(B\cap C)=P(B)P(C)=\frac14$. – hmakholm left over Monica Aug 19 '17 at 20:42
• I don't see it that simple. Can you explain me? – Bren Amador Aug 19 '17 at 20:45
• x @Bren: What is your problem with this? Do you not agree that the definition of "$B$ and $C$ are independent" is "$P(B\cap C)=P(B)P(C)$"? Do you not agree that $P(B\cap C)=\frac 14$? Do you not agree that $P(B)P(C)=\frac 14$? Do you not agree that probabilities that both equal $\frac14$ will also equal each other? Give me something to work with here. – hmakholm left over Monica Aug 19 '17 at 20:49
• I'm guessing these are regular six sided dice ? – user451844 Aug 19 '17 at 21:02

Thus $(A \cap C) = (A\cap B^\complement)$ and $(B\cap C)=(B\cap A^\complement)$.
Hint: $\mathsf P(A) = \mathsf P(A\cap B)+\mathsf P(A\cap B^\complement)$ by the Law of Total Probability