Refer to Example 13. (i) Complete the following table: Event : ‘Sum on 2 dice’ 2 3 4 5 6 7 8 9 10 11 12 Probability 1 36 5 36 1 36 (ii) A student argues that ‘there are 11 possible outcomes 2, 3, 4, 5, 6, 7, 8, 9, 10, 11 and
Step-by-Step Solution
Key Concept: The probability of an event = (Number of favourable outcomes) / (Total number of equally likely outcomes). For the sum of two dice, the total number of equally likely outcomes is $6 \times 6 = 36$. Different sums have different numbers of favourable ordered pairs, so their probabilities are not equal.
### (i) Completing the table
1. List all ordered pairs (die‑1, die‑2) that give each possible sum.
- Sum = 2 : (1,1) → 1 way
- Sum = 3 : (1,2), (2,1) → 2 ways
- Sum = 4 : (1,3), (2,2), (3,1) → 3 ways
- Sum = 5 : (1,4), (2,3), (3,2), (4,1) → 4 ways
- Sum = 6 : (1,5), (2,4), (3,3), (4,2), (5,1) → 5 ways
- Sum = 7 : (1,6), (2,5), (3,4), (4,3), (5,2), (6,1) → 6 ways
- Sum = 8 : (2,6), (3,5), (4,4), (5,3), (6,2) → 5 ways
- Sum = 9 : (3,6), (4,5), (5,4), (6,3) → 4 ways
- Sum = 10 : (4,6), (5,5), (6,4) → 3 ways
- Sum = 11 : (5,6), (6,5) → 2 ways
- Sum = 12 : (6,6) → 1 way
2. Compute probability for each sum using the formula
$$P(\text{sum}=s) = \frac{\text{Number of favourable ordered pairs}}{36}$$
3. Fill the table
| Sum (s) | Favourable outcomes | Probability |
|---------|---------------------|-------------|
| 2 | 1 | $\frac{1}{36}$ |
| 3 | 2 | $\frac{2}{36}$ |
| 4 | 3 | $\frac{3}{36}$ |
| 5 | 4 | $\frac{4}{36}$ |
| 6 | 5 | $\frac{5}{36}$ |
| 7 | 6 | $\frac{6}{36}$ |
| 8 | 5 | $\frac{5}{36}$ |
| 9 | 4 | $\frac{4}{36}$ |
| 10 | 3 | $\frac{3}{36}$ |
| 11 | 2 | $\frac{2}{36}$ |
| 12 | 1 | $\frac{1}{36}$ |
### (ii) Addressing the student’s argument
- The student is correct that there are 11 distinct sums (2 to 12). However, the statement *“there are 11 possible outcomes”* is misleading when used for probability calculation.
- In probability, we need the number of equally likely elementary outcomes. For two dice, the elementary outcomes are the ordered pairs, and there are $6 \times 6 = 36$ of them, not 11.
- Because the 11 sums are not equally likely, each sum has a different probability as shown in the completed table above. Hence, we cannot assign a probability of $\frac{1}{11}$ to each sum.
- The correct approach is to count the favourable ordered pairs for each sum and divide by 36, as demonstrated in part (i).
Correct Answer: Completed table:
| Sum | Probability |
|-----|-------------|
| 2 | $\frac{1}{36}$ |
| 3 | $\frac{2}{36}$ |
| 4 | $\frac{3}{36}$ |
| 5 | $\frac{4}{36}$ |
| 6 | $\frac{5}{36}$ |
| 7 | $\frac{6}{36}$ |
| 8 | $\frac{5}{36}$ |
| 9 | $\frac{4}{36}$ |
| 10 | $\frac{3}{36}$ |
| 11 | $\frac{2}{36}$ |
| 12 | $\frac{1}{36}$ |
Explanation for part (ii): The student’s claim that there are 11 possible outcomes is true only for the *values* of the sum, but the elementary outcomes are 36 ordered pairs. Therefore, probabilities must be calculated using 36 as the denominator, not 11.