Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2014 / Paper 1 / 18 / c

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

i

Solution

The total number of two-fish hauls is

(2 10) = 45

. Selecting both from the eight carp gives

P (X = 0) = \frac{( 0 2 ) ( 2 8 )}{45} = \frac{28}{45} .

ii

Solution

Choose one of the two trout and one of the eight carp. The hypergeometric distribution gives

P (X = 1) = \frac{( 1 2 ) ( 1 8 )}{45} = \frac{16}{45} .

iii

Solution

Only one labelled haul contains both trout, giving

P (X = 2) = \frac{( 2 2 ) ( 0 8 )}{45} = \frac{1}{45} .

Solution

The three preceding probabilities satisfy

\frac{28}{45} + \frac{16}{45} + \frac{1}{45} = 1.

Their expected value and second moment are

μ = \frac{16 + 2}{45} = \frac{2}{5}, E [X^{2}] = \frac{16 + 4}{45} = \frac{4}{9} .

Subtracting the squared mean gives

μ = \frac{2}{5}, σ^{2} = \frac{4}{9} - \frac{4}{25} = \frac{64}{225} .

The value

μ = 2 (2/10)

also follows from linearity of expectation. The variance includes the finite-population reduction caused by sampling without replacement.

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