Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2012 / Paper 1 / 18 / a

Under independent daily tosses, attending every lecture is the intersection of

24

3/4

. Hence

P (N = 24) = (\frac{3}{4})^{24} \approx 0.00100339.

ii

Attending none requires a tail on every independent toss. Thus

P (N = 0) = (\frac{1}{4})^{24} = 2^{- 48} \approx 3.55271 \times 1 0^{- 15} .

Write the count as a sum of

24

3/4

E [N] = 24 \cdot \frac{3}{4} = 18 lectures .

Equivalently,

N

24

and

3/4

Independence makes the variances of the attendance indicators additive. The binomial distribution therefore gives

Var (N) = n p (1 - p) = 24 (3/4) (1/4) = 9/2

σ_{N} = \frac{3}{2} \approx 2.12132 lectures .

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