I'm taking stats course and currently, I'm studying basic probability on my own.
There's this one sample question that I still can't solve so far and I need your help.
Here's the question:
A random experiment of rolling a 6-sided die twice.
Let A be the event of getting same number on both rolls, B the event that the rolls add up to 4, and C the event that the rolls add up to 2. Compute P(Ac), P(A ∩ B), P(A ∪ B), P(B ∩ C).
The problems is i get different answers for P(A ∪ B), P(B ∩ C)
(According to my instructor, the answers for them are 2/9 and 0 respectively.)
Here's how i solved them:
1. For P(A ∪ B):
A = the event of getting same number on both rolls
A = { {1, 1}, {2, 2}, {3, 3}, {4, 4}, {5, 5}, {6, 6} }
P(A) = number of outcomes of A / number of All outcomes = 6/36 = 1/6
B = the event that the rolls add up to 4
B = { {1, 1}, {1, 2}, {1, 3}, {2, 1}, {2, 2}, {3, 1} }
P(B) = number of outcomes of B / number of All outcomes = 6/36 = 1/6
P(A ∩ B) = P(B | A) * P(A)
but P(B | A) = P(B) therefore
P(A ∩ B) = P(B) * P(A) = 1/6 * 1/6 = 1/36
P(A ∪ B) = P(A) + P(B) - P(A ∩ B)
since P(A ∩ B) is not ∅
P(A ∪ B) = P(A) + P(B) - P(A ∩ B) = 1/6 + 1/6 - 1/36 = 6/36 + 6/36 - 1/36 = 11/36
P(A ∪ B) = 11/36
2. For P(B ∩ C):
P(B) = 1/6
C = the event that the rolls add up to 2
C = { {1, 1} }
P(C) = 1/36
P(B ∩ C) = P(B) * P(C) since P(B|C) = P(B)
therefore
P(B ∩ C) = 1/6 * 1/36 = 1/216
Please correct me if any part of the solution is wrong.
Thank you!
The question states that B is the event that the dice rolls add up to exactly 4 not less than or equal to 4. So:
B = {{1, 3}, {2, 2}, {3, 1}}
P(A ∪ B) = |(A ∪ B)|/36 = (|A| + |B| - |(A ∩ B)|)/36 = 8/36 = 2/9
P(B ∩ C) = |(B ∩ C)|/36 = 0