 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2025/paper-1/10/b/solution

The favourable outcomes are

(j, j + 1)

and

(j + 1, j)

for

j = 1, \dots, 5

. These are ten distinct ordered outcomes among the

36

equally likely elements of the sample space, so the probability is

\frac{10}{36} = \frac{5}{18} .

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