Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2005 / Paper 1 / 3 / c / Solution

The quoted pass probabilities on later attempts are conditional on reaching those attempts. If

K

is the first successful attempt, then

P (K = 1) = \frac{3}{5}, P (K = 2) = \frac{2}{5} \frac{4}{5} = \frac{8}{25}, P (K = 3) = \frac{2}{5} \frac{1}{5} \frac{7}{8} = \frac{7}{100} .

They sum to

99/100

, so conditioning on success by the third attempt is essential. The conditional mean of a bounded success time is

E [K ∣ K \leq 3] = \frac{1 ( 3/5 ) + 2 ( 8/25 ) + 3 ( 7/100 )}{99/100} = \frac{145}{99} ≃ 1.465.

The remaining probability

1/100

represents failing all three attempts and is excluded by this conditioning.

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