Question

during a charity las vegas casino night, rosie plays craps and gets to roll the dice. use the sample space for rolling two dice as shown below. express your answers as reduced fractions.
find the probability the roll was a
(a) sum of 12.
the probability the roll was a sum of 12 is
(b) sum that is even.
the probability the roll was a sum that is even is

Explanation:

Step1: Count total number of outcomes

When rolling two dice, the total number of outcomes is \(n(S)=36\) (as shown in the sample - space grid).

Step2: Count number of even - sum outcomes

Let's consider the sum of two numbers \(a + b\) (where \(a\) is the outcome of the first die and \(b\) is the outcome of the second die). The sum \(a + b\) is even if either both \(a\) and \(b\) are even or both \(a\) and \(b\) are odd.

• Number of ways to get both odd: There are 3 odd numbers (\(1,3,5\)) on a die. The number of ways to get an odd number on the first die and an odd number on the second die is \(n_{1}=3\times3 = 9\) (e.g., \((1,1),(1,3),(1,5),(3,1),(3,3),(3,5),(5,1),(5,3),(5,5)\))
• Number of ways to get both even: There are 3 even numbers (\(2,4,6\)) on a die. The number of ways to get an even number on the first die and an even number on the second die is \(n_{2}=3\times3=9\) (e.g., \((2,2),(2,4),(2,6),(4,2),(4,4),(4,6),(6,2),(6,4),(6,6)\))

The number of favorable outcomes \(n(A)=n_{1}+n_{2}=18\)

Step3: Calculate the probability

The probability formula is \(P(A)=\frac{n(A)}{n(S)}\). Substituting \(n(A) = 18\) and \(n(S)=36\) into the formula, we get \(P(A)=\frac{18}{36}=\frac{1}{2}\)

Answer:

\(\frac{1}{2}\)