In a simultaneous throw of a pair of dice, find the probability of getting:
neither 9 nor 11 as the sum of the numbers on the faces.

Answer

To find the probability of getting neither 9 nor 11 as the sum of the numbers on the faces when throwing a pair of dice, we can follow these steps:

### Step 1: Determine the Total Outcomes
When two dice are thrown, each die has 6 faces. Therefore, the total number of outcomes when throwing two dice is:
\[ \text{Total Outcomes} = 6 \times 6 = 36 \]

### Step 2: Find the Outcomes for Sum of 9
Next, we need to find the combinations that result in a sum of 9. The pairs that give a sum of 9 are:
- (3, 6)
- (4, 5)
- (5, 4)
- (6, 3)

Thus, there are 4 outcomes that result in a sum of 9.

### Step 3: Find the Outcomes for Sum of 11
Now, we find the combinations that result in a sum of 11. The pairs that give a sum of 11 are:
- (5, 6)
- (6, 5)

Thus, there are 2 outcomes that result in a sum of 11.

### Step 4: Calculate the Total Outcomes for Sums of 9 and 11
Now, we add the outcomes for sums of 9 and 11:
\[ \text{Total Outcomes for 9 and 11} = 4 + 2 = 6 \]

### Step 5: Calculate the Outcomes that are Neither 9 nor 11
To find the outcomes that are neither 9 nor 11, we subtract the outcomes for 9 and 11 from the total outcomes:
\[ \text{Outcomes neither 9 nor 11} = 36 - 6 = 30 \]

### Step 6: Calculate the Probability
Finally, the probability of getting neither 9 nor 11 is given by the ratio of favorable outcomes to total outcomes:
\[ \text{Probability} = \frac{\text{Favorable Outcomes}}{\text{Total Outcomes}} = \frac{30}{36} = \frac{5}{6} \]

Thus, the required probability of getting neither 9 nor 11 is:
\[ \frac{5}{6} \]