Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2025 / Paper 2 / 12 / e

Table of contents
- i e
  - Solution i
- ii e
  - Solution ii
- iii e
  - Solution iii

i

Solution

Only a green ball carries label four. After it is removed, nine of the remaining nineteen balls are green. The conditional probability is

\frac{9}{19} .

ii

Solution

The specified colour tells us that one green ball has been removed. Thus sampling without replacement again leaves nine green balls among nineteen:

\frac{9}{19} .

The numerical label supplies no further change once this first colour is known.

iii

Solution

There are two equally likely balls bearing label eight, one of each colour. The conditional probability that the removed ball was green is therefore

1/2

. By the law of total probability,

P (second green ∣ X_{1} = 8) = \frac{1}{2} \frac{9}{19} + \frac{1}{2} \frac{10}{19} = \frac{1}{2} .

Both possible first colours must be included because the observed label alone does not determine the colour.

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