(i) A pair of dice is rolled. Find the probability of getting a doublet or sum of numbers to be at least 10.
(ii) Two dice are tossed together. Find the probability of getting a doublet or (a) total o 10 (b) total of 6.

Answer

To solve the given problems, we will follow these steps:

### Part (i): A pair of dice is rolled. Find the probability of getting a doublet or sum of numbers to be at least 10.

1. **Identify the Total Outcomes**:
When a pair of dice is rolled, each die has 6 faces. Therefore, the total number of outcomes when rolling two dice is:
\[ \text{Total Outcomes} = 6 \times 6 = 36 \]

2. **Identify the Favorable Outcomes**:
- **Doublets**: A doublet occurs when both dice show the same number. The doublets possible are (1,1), (2,2), (3,3), (4,4), (5,5), and (6,6). Thus, there are 6 doublets.
- **Sum of Numbers at Least 10**: The combinations that yield a sum of at least 10 are:
- (4,6), (5,5), (6,4), (5,6), (6,5), (6,6)
- The pairs that give a sum of 10 or more are: (4,6), (5,5), (6,4), (5,6), (6,5), (6,6).
- The outcomes are: (4,6), (5,5), (6,4), (5,6), (6,5), (6,6) which gives us 6 outcomes.

However, we have already counted (5,5) and (6,6) in both categories. Hence, we need to avoid double counting.

- **Total Favorable Outcomes**:
\[ \text{Total Favorable Outcomes} = \text{Doublets} + \text{Sum at least 10} - \text{Doublets counted in Sum at least 10} \]
\[ = 6 + 6 - 2 = 10 \]

3. **Calculate the Probability**:
The probability \( P \) of getting a doublet or a sum of numbers at least 10 is given by:
\[ P = \frac{\text{Total Favorable Outcomes}}{\text{Total Outcomes}} = \frac{10}{36} = \frac{5}{18} \]

### Part (ii): Two dice are tossed together. Find the probability of getting a doublet or:
(a) total of 10
(b) total of 6.

1. **For Total of 10**:
- The combinations that yield a sum of 10 are: (4,6), (5,5), (6,4).
- The total outcomes that give a sum of 10 are 3.
- Since there is no overlap with doublets (only (5,5) is a doublet), we have:
\[ \text{Total Favorable Outcomes for Total of 10} = 3 \text{ (including doublet (5,5))} \]
\[ P(\text{Total of 10}) = \frac{3}{36} = \frac{1}{12} \]

2. **For Total of 6**:
- The combinations that yield a sum of 6 are: (1,5), (2,4), (3,3), (4,2), (5,1).
- The total outcomes that give a sum of 6 are 5.
- Again, (3,3) is a doublet, so:
\[ \text{Total Favorable Outcomes for Total of 6} = 5 \text{ (including doublet (3,3))} \]
\[ P(\text{Total of 6}) = \frac{5}{36} \]

### Summary of Probabilities:
- (i) Probability of getting a doublet or sum of numbers at least 10: \( \frac{5}{18} \)
- (ii)(a) Probability of getting a doublet or total of 10: \( \frac{1}{12} \)
- (ii)(b) Probability of getting a doublet or total of 6: \( \frac{5}{36} \)