# Probability Independence with two dice

1. Two dice, one red and one green, are rolled. Define the events A: the red die lands on a 3, 4, or 5 B: the sum of the two dice is 9 (a) Compute P(A | B). (b) Compute P(B | A). (c) Are A and B independent events? Justify your answer.

So I'm kinda stuck. I have:

P(A)=3/6=1/2

P(B)=(6-|7-9|)/36=1/9

Though, I can't remember how to calculate P(A|B) and P(B|A), I know there's the Bayes theoreom, but I don't know P(B|A) so I can't calculate P(A|B).

• Bayes' theorem will give you each answer individually. It doesn't require knowing $P(B|A)$ to calculate $P(A|B)$. But you will need to know $P(A \land B)$. Oct 12 at 2:41
• When you calculated P(A), the denominator of 6 was the number of all possible outcomes, and the numerator 3 came from the 3 numbers described by A. Now if you are given A conditionally, that event becomes your denominator. So instead of 6 possible outcomes in the denominator, there are only 3, either the die is 3 or it's 4 or it's 5. If you want to know P(B|A), all possibilities are 3, and P(B) is the sum being 9 when one die is 3, 4, or 5. You can solve it the same way as P(A), by the same logic, or else employ the formula for conditional probability following @aschepler Oct 12 at 2:59

You also need to find the probability of the intersection event $$\operatorname{P}(A\cap B)$$. Note that in this exercise $$A\cap B=\{\text{the red die rolls 3 or 4 or 5 AND the sum on the two dice is 9}\}$$, and so its probability is …
Then you can use the formula for (which, in fact, is the definition of) conditional probabilities: $$\operatorname{P}(A\mid B)=\frac{\operatorname{P}(A\cap B)}{\operatorname{P}(B)}$$.
To determine whether the two events are independent, use the fact that two events $$A$$ and $$B$$ with nonzero probabilities are independent if and only if $$\operatorname{P}(A\cap B)=\operatorname{P}(A)\cdot\operatorname{P}(B)$$.