Question

1. Two dice are rolled. a) Define the sample space for this probability experiment. How many outcomes are there? What are the possible outcomes? b) Find the probability of each of the following outcomes: i) Snake Eyes (2 one’s) ii) A sum greater than 4 iii) An even sum iv) At least one die is a 3 v) An odd sum or a sum of 8

Answer 1

a) First die can take any number from 1 t 6, giving 6 possibilities.
2nd die can also take 6 possibilities. So in total, there are 6 * 6 = 36 outcomes.
The possible outcomes are:
S = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6),
(2, 1), ... (2, 6),
...
(6, 1) ... (6, 6)}

Answer 2

i) Since there is only one way to get Snake eyes (1,1), the probability is 1/36

ii) A sum greater than 4 means you have either 5 or more. This will gives lots of possibilities, so consider the complement event (A sum of 4 or less). You can get 4 or less this way:
(1, 1), (1, 2), (1, 3),
(2, 1), (2, 2)
(3, 1)

So there are 6 outcomes. Since we looked at the complement, our event of interest has 36 - 6 = 30 outcomes. SO the probability is 30/36

iii) An even sum means you can get:
(1, 1), (1, 3), (1, 5)
(2, 2), (2, 4), (2, 6)
...
(6, 2), (6, 4), (6, 6)

This gives 18 outcomes, so the probability is 18/36

iv) At least one die is 3. That means you can get one die is 3, or both are 3, meaning you can get these possibilities:
(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6)
(1, 3), (2, 3), , (4, 3), (5, 3), (6, 3)

There are 11 such possibilities, giving a probability of 11/36 (notice that I skipped (3,3) in the second row, because we already have it in the first row, and you don't want to double count).

v) To get an odd sum... well we found that the probability of an even sum was 18/36, so the probability for an odd sum is also 18/36.
To get a sum of 8, you can do it with these possibilities:
(2, 6), (3, 5), (4, 4), (5, 3), (6, 2)

Then, the probability to get an odd sum OR a sum of 8 is the union of these events. If
\(A\)= odd sum
\(B\) = sum is 8

Then \(P(A \cup B) = P(A)+P(B)-P(A\cap B)\), but for the intersection event, you don't have to worry about it since the probability of the sum being odd and 8 is not possible, so A and B are mutually exclusive so \(P(A \cap B)=0\)