Past exam of the mathematics course of the University of Cambridge / 2011 / ia / Paper 2 / 12F / b / iii

Table of contents
- Solution iii

Solution

The event that

H_{1}

is hit but

H_{2}

is not has probability

1/3 - 1/12 = 1/4

. Dividing by the probability

1/3

that

H_{1}

is covered, using the definition of conditional probability, gives

P (H_{2} not hit ∣ H_{1} hit) = \frac{1/3 - 1/12}{1/3} = \frac{3}{4} .

This conditioning concerns angular coverage of a sector; the two house-hit events are not independent.

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