Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2019 / Paper 1 / 16 / b

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

i

Solution

For sampling without replacement, sum the two outcomes with a blue second draw:

P (B_{2}) = \frac{1}{10} + \frac{3}{10} = \frac{2}{5} (r = 0) .

The marginal blue probability equals the initial blue fraction even though the two draws are dependent.

ii

Solution

For sampling with replacement, the box has its original contents at the second draw. Therefore

P (B_{2}) = \frac{2}{5} (r = 1) .

iii

Solution

Condition on whether replacement occurs. Both alternatives have blue second-draw probability

2/5

, so the law of total probability gives

P (B_{2}) = r \frac{2}{5} + (1 - r) \frac{2}{5} = \frac{2}{5}, 0 \leq r \leq 1.

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