Dice probability

[heathend0]

I'm unsure if I completed the following probability correctly, I included my answer below and I

Consider the following model, which corresponds to repeatedly rolling two dice d1 and d2 and stopping the first time d1+d2∈{5,7}.
You roll two 6-sided dice d1 and d2. Consider the events
A = “d1+d2=7”
B = “d1+d2=5”
C=A∪B

1. What is Pr(A∣C)?

my answer:
P(A) = 6/36 = 1/6
P(B) = 4/36 = 1/9
P(C) = P(A∪B) = P(A) + P(B) since A and B are mutually exclusive = 5/18

P(A|C) = P(A and C) / P(C)
A and C can only be true when A is true, so P(A and C) = P(A)
1/6 ÷ 5/18 = 3/5

 

[Steven G]

Your answers are all correct. Great job.

Here is another way of thinking about P(A|C)

If we restrict our sample space to C then there are 10 outcomes. 6 out of these 10 outcomes are in set A. So P(A|C) = 6/10 or 3/5

 

[pka]

I agree that your answers are correct.
One note: There are six pairs in event \(A\) and four pairs in event \(B\).
We know that \(A\cap B=\emptyset\), thus event \(C\) has ten pairs.
Thus \(\mathcal{P}(A|C)\) is simply \(\dfrac{6}{10}\).

 

[heathend0]

Great thanks guys!