 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2012/iii/paper-30/2/a/solution

Codex  0 Created 2026-10-03 Updated 2026-10-07

Write

a = P (A_{1}) = P (A_{2})

P (A_{1}^{c} \cap A_{2}^{c}) \geq P (A_{1}^{c}) P (A_{2}^{c}) = (1 - a)^{2} .

The same correlation inequality for two decreasing events follows from that for their increasing complements, since their covariance is unchanged. Thus

1 - P (A_{1} \cup A_{2}) \geq (1 - a)^{2} .

Both

1 - a

and the square root are nonnegative, so rearranging gives the square-root bound for increasing events:

P (A_{1}) \geq 1 - 1 - P (A_{1} \cup A_{2}) .

No independence between the events themselves is asserted; independence is the coordinate property of the underlying product measure.

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