To calculate the theoretical probability of rolling a total of 7 or 11 with a pair of number cubes (dice), we first need to determine the total number of possible outcomes when two dice are rolled and then the number of favorable outcomes that result in either a total of 7 or 11.
Total Outcomes
When rolling a pair of six-sided dice, the total number of outcomes is: \[ 6 \text{ (sides of die 1)} \times 6 \text{ (sides of die 2)} = 36. \]
Favorable Outcomes for a Total of 7
The combinations that yield a total of 7 are:
- (1, 6)
- (2, 5)
- (3, 4)
- (4, 3)
- (5, 2)
- (6, 1)
This gives us a total of 6 favorable outcomes for rolling a total of 7.
Favorable Outcomes for a Total of 11
The combinations that yield a total of 11 are:
- (5, 6)
- (6, 5)
This gives us a total of 2 favorable outcomes for rolling a total of 11.
Total Favorable Outcomes
Now, we add the favorable outcomes for both totals: \[ 6 \text{ (for 7)} + 2 \text{ (for 11)} = 8 \text{ favorable outcomes.} \]
Probability Calculation
The probability \( P \) of rolling a total of 7 or 11 is given by the ratio of the number of favorable outcomes to the total number of outcomes: \[ P(7 \text{ or } 11) = \frac{\text{Number of favorable outcomes}}{\text{Total outcomes}} = \frac{8}{36}. \]
Simplifying the Fraction
To simplify \( \frac{8}{36} \), we divide both the numerator and the denominator by their greatest common divisor (GCD), which is 4: \[ \frac{8 \div 4}{36 \div 4} = \frac{2}{9}. \]
Final Answer
Thus, the theoretical probability of rolling a total of 7 or 11 on a pair of number cubes from a single toss is: \[ P(7 \text{ or } 11) = \frac{2}{9}. \]