To solve the problem, we first need to identify all possible outcomes when rolling the cube twice and then find the pairs of outcomes that sum to 4.
The number cube has the following sides: 1, 1, 2, 2, 3, 3.
When Sam rolls the cube twice, we can represent the possible outcomes as pairs (first roll, second roll).
Possible outcomes:
- (1, 1)
- (1, 2)
- (1, 3)
- (1, 1)
- (1, 2)
- (1, 3)
- (2, 1)
- (2, 2)
- (2, 3)
- (2, 1)
- (2, 2)
- (2, 3)
- (3, 1)
- (3, 2)
- (3, 3)
- (3, 1)
- (3, 2)
- (3, 3)
(Note that some pairs are repeated because of the duplicate numbers on the cube.)
Next, we will calculate the total number of combinations we can have when rolling the cube twice:
- From the first roll: 1, 1, 2, 2, 3, 3 (6 outcomes)
- From the second roll: 1, 1, 2, 2, 3, 3 (6 outcomes)
Thus, the total number of outcomes when rolling the cube twice is: \[ 6 \times 6 = 36 \]
Now, let's find the outcomes where the sum is 4:
We will list the pairs that sum to 4:
- (1, 3)
- (3, 1)
- (2, 2)
Now, we will count the number of ways each pair can occur based on the cube's sides.
-
For (1, 3):
- The 1 can come from either of the two 1's.
- The 3 can come from either of the two 3's.
- Ways: 2 choices for 1 and 2 choices for 3 → \(2 \times 2 = 4\).
-
For (3, 1):
- The 3 can come from either of the two 3's.
- The 1 can come from either of the two 1's.
- Ways: 2 choices for 3 and 2 choices for 1 → \(2 \times 2 = 4\).
-
For (2, 2):
- Both 2's can match up, resulting in only one way: (2, 2) → 2 choices for 2 and 2 choices for 2 → \(2 \times 2 = 4\).
Adding them together:
- Ways for (1, 3) = 4
- Ways for (3, 1) = 4
- Ways for (2, 2) = 4
Total ways to achieve a sum of 4 = \(4 + 4 + 4 = 12\).
Now, we calculate the probability:
The probability \(P\) that the sum of the numbers will add to 4 is given by \[ P = \frac{\text{Number of favorable outcomes}}{\text{Total outcomes}} = \frac{12}{36} = \frac{1}{3} \]
Thus, the probability that the sum of the numbers rolled will equal 4 is:
\(\boxed{\frac{1}{3}}\)
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