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Let $\Omega = \{(i,j):i,j=1,...,6\}, A= \{(i,j):i=1,3,5\}$, and $B=\{(i,j): i+j=8\}$. Find $P(A|B)$. Are $A$ and $B$ independent events?

$\Omega = \{1,2,3,4,5,6\}$, Then $P(A)= \frac{1}{2}$, and $B=(2,6),(3,5),(4,4),(5,3),(6,2)$ so $P(B)=\frac{1}{3}$

Then $P(A|B)=\frac{P(A\cap B)}{P(B)}= \frac{\frac{1}{2}\cdot\frac{5}{36}}{\frac{5}{36}}=\frac{1}{2}$

Am I on the right direction?

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    No. The $\Omega$ in the question formulation contains $36$ elements, not $6$. However, $P(A)$ is still correct. Everything concerning $B$ is a mess. – 2017-02-16
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    Why is it 36? and B has the parameters of $i+j=8$ – 2017-02-16
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    Because $6\times 6 = 36$. – 2017-02-16
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    Oh okay, so then B can be $\{7,5,3\}?$ – 2017-02-16
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    No, it can't. $B$, as a subset of $\Omega$, should consist of *pairs* (of numbers that add up to $8$). – 2017-02-16
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    So $B:(i+j)=\{(7,1),(5,3), (5,3)\}$ since A only has $\{1,3,5\}$ – 2017-02-16
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    Not even close. Look at the definition of $B$. Do you see $A$ somewhere? – 2017-02-16
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    Oh i see my mistake I was assuming $A=i$ Then $P(B) =\{1,7\},\{2,6\},\{3,5\},\{4,4\},\{5,3\},\{6,2\},\{7,1\}$? – 2017-02-16
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    It's $B$, not $P(B)$. – 2017-02-16
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    But is the set correct? – 2017-02-16
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    Well, almost. Since, as I've said, it should be a subset of $\Omega$, $7$ cannot be there. – 2017-02-16
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    Then I just eliminate $\{7,1\}$ and vice versa? – 2017-02-16
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    Yes. (Btw I also wonder why parentheses became braces.) – 2017-02-16
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    Then P(B)=$\frac{5}{36}$ – 2017-02-16
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    This is correct. – 2017-02-16
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    Oh okay, then I got the conditional part now, but how can I prove they are independent? – 2017-02-16
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    I don't see where you got the conditional part. Your formula for $P(A\mid B)$ is some nonsense. – 2017-02-16
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    Isn't $P(A|B) = \frac{P(A \cap B)}{P(B)}?$ – 2017-02-16
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    Yes, but it is not what is written in your post. – 2017-02-16
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    I haven't updated it yet, but i will right now – 2017-02-16
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    How can i show both events are independent? – 2017-02-16
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    $A$ and $B$ aren't independent. Consider the event $A\cap B$. It consists of those elements of $B$ whose first component is one of $1,3$, and $5$. Thus how many elements does $A\cap B$ have? Using that, what is $P(A\mid B) = \dfrac{P(A\cap B)}{P(B)}$? – 2017-02-16
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    Would it be the P(A|B) I have on the post? – 2017-02-16

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After more than 20 Comments, perhaps it is worthwhile to put the pieces together.

In plain English, you roll two ordinary (six-sided) fair dice independently. One die is red and the other is green. If the total on the two dice is 8, then what is the probability the red die showed an odd number?

Of the five sample points $(i,j)$ listed in which the total is eight, only two of them have an odd first element ($i$). So the answer is $2/5 = 0.4.$

In the equation $P(A|B) = P(A \cap B)/P(B),$ we have $P(B) = 5/36$ and $P(A \cap B) = 2/36.$ So $P(A|B) = \frac{2/36}{5/36} = 2/5.$

Note: Events $A$ and $B$ are not independent, so you can't write $P(A \cap B) = P(A)P(B).$ In fact, $P(A) = 1/2$ and $P(B) = 5/36,$ whereas $P(A \cap B) = 2/36.$