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:



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).

  • $\begingroup$ 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 $P(A \land B)$. $\endgroup$
    – aschepler
    Oct 12 at 2:41
  • $\begingroup$ 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 $\endgroup$ Oct 12 at 2:59

The sum of the dice can be 9 only when the red die shows 3, 4, 5, or 6.

When the red die does show one of these faces, there is only one face for each that the green die may show to make the sum 9.


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 $\operatorname{P}(A\cap B)=\operatorname{P}(A)\cdot\operatorname{P}(B)$.


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