Science:Math Exam Resources/Courses/MATH103/April 2012/Question 02 (e)/Solution 1

This game has 9 outcomes, but due to symmetry only 6 of them are distinct. The probabilities are distributed as follows (multiples of 1/64)

	pig 1 side	pig 1 feet	pig 1 snout
pig 2 side	4 × 4	4 × 3	4 × 1
pig 2 feet	3 × 4	3 × 3	3 × 1
pig 2 snout	1 × 4	1 × 3	1 × 1

Let X be the random variable that returns the score. Then

${\begin{aligned}p(X=0)&={\frac {16}{64}}&p(X=1)&=2{\frac {12}{64}}&p(X=5)&=2{\frac {4}{64}}\\p(X=3)&={\frac {9}{64}}&p(X=7)&=2{\frac {3}{64}}&p(X=10)&={\frac {1}{64}}\end{aligned}}$

and thus the average $\mu$ is found to be

${\begin{aligned}\mu &=0p(X=0)+1p(X=1)+3p(X=3)+5p(X=5)+7p(X=7)+10p(X=10)\\&=2{\frac {12}{64}}+3{\frac {9}{64}}+10{\frac {4}{64}}+14{\frac {3}{64}}+10{\frac {1}{64}}\quad ={\frac {143}{64}}\approx 2.23\end{aligned}}$