Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2006 / Paper 2 / 8

Assuming the egg colours are independent with the same probability $p$, the number of white eggs has the binomial distribution with four trials. The possibilities are

The binomial coefficients count which eggs are white. Marginal colour probabilities alone would not determine nest probabilities without the independence assumption.

For $p=3/4$, the five probabilities in increasing white-egg order are $(1,12,54,108,81)/256$. The largest is $108/256=27/64$, so the most common nest has three white eggs and one brown egg, $W_3{B}_1$.

The expected value of the discrete uniform distribution is

Using the sum of squares, the second moment is

The variance is the second moment minus the square of the expected value:

The function is a probability density function, rather than a cumulative distribution: its integral over both intervals must be one. Thus $2\alpha +2\alpha =1$ and

The density is symmetric about $7/2$, giving expected value $\mathbb{E}[X]=7/2$. Its second moment is $\mathbb{E}[X^2]=\frac{1}{4}({\int }_1^3{x}^{2}dx+{\int }_4^6{x}^{2}dx)=89/6$. ThereforeEquivalently, each equally weighted interval has variance $1/3$ and its centre is $3/2$ from the overall mean, giving $1/3+9/4$ by the law of total variance.