Dice Probability: Sum Neither 9 Nor 10

This problem asks for the probability of an event occurring when two perfect dice are thrown. Specifically, we need to find the chances that the sum of the numbers shown on the two dice is not equal to 9 and also not equal to 10.

Understanding the Sample Space

When two perfect six-sided dice are thrown, each die can land on any integer from 1 to 6. The total number of possible outcomes is the product of the number of outcomes for each die.

Total possible outcomes = (Number of faces on Die 1) \(\times\) (Number of faces on Die 2)

Total possible outcomes = \(6 \times 6 = 36\).

These 36 outcomes form the sample space. Each outcome is equally likely.

Calculating Outcomes for Specific Sums

We are interested in the sums 9 and 10. Let's list the combinations of numbers from the two dice that result in these sums:

Outcomes for Sum = 9

The pairs of numbers that add up to 9 are:

• (3, 6)
• (4, 5)
• (5, 4)
• (6, 3)

There are 4 outcomes where the sum is 9.

Outcomes for Sum = 10

The pairs of numbers that add up to 10 are:

• (4, 6)
• (5, 5)
• (6, 4)

There are 3 outcomes where the sum is 10.

Probability of Sum Being 9 or 10

The event "sum is 9" and the event "sum is 10" are mutually exclusive, meaning they cannot happen at the same time. Therefore, the number of outcomes where the sum is either 9 or 10 is the sum of the outcomes for each case.

Number of outcomes for sum 9 or 10 = (Outcomes for sum 9) + (Outcomes for sum 10)

Number of outcomes for sum 9 or 10 = \(4 + 3 = 7\).

The probability of the sum being 9 or 10 is calculated as:

\(P(\text{Sum}=9 \text{ or } \text{Sum}=10) = \frac{\text{Number of outcomes for sum 9 or 10}}{\text{Total number of outcomes}}\)

\(P(\text{Sum}=9 \text{ or } \text{Sum}=10) = \frac{7}{36}\)

Probability of Sum Being Neither 9 Nor 10

The question asks for the probability that the sum is *neither* 9 *nor* 10. This is the complement of the event that the sum *is* 9 or 10.

The probability of a complement event is found by subtracting the probability of the event from 1.

\(P(\text{Sum} \neq 9 \text{ and } \text{Sum} \neq 10) = 1 - P(\text{Sum}=9 \text{ or } \text{Sum}=10)\)

\(P(\text{Sum} \neq 9 \text{ and } \text{Sum} \neq 10) = 1 - \frac{7}{36}\)

To subtract, we find a common denominator:

\(1 = \frac{36}{36}\)

\(P(\text{Sum} \neq 9 \text{ and } \text{Sum} \neq 10) = \frac{36}{36} - \frac{7}{36}\)

\(P(\text{Sum} \neq 9 \text{ and } \text{Sum} \neq 10) = \frac{36 - 7}{36}\)

\(P(\text{Sum} \neq 9 \text{ and } \text{Sum} \neq 10) = \frac{29}{36}\)

Therefore, the probability that the sum of the numbers on the faces is neither 9 nor 10 is \(\frac{29}{36}\).