Question

The following three games are scheduled to be played at the World Curling Championship one morning. The values in parentheses are the probabilities of each team winning their respective game.
Game 1: Finland (0.2) vs. Canada (0.8)

Game 2: USA (0.3) vs. Switzerland (0.7)

Game 3: Germany (0.4) vs. Japan (0.6)
(a) The outcome of interest is the set of winners for each of the three games. List the complete sample space of outcomes and calculate the probability of each outcome.

(b) Let X be the number of European teams that win their respective games. Find the probability distribution of X.

(c) Find the expected value and variance of X.

(d) If two European teams win their games, what is the probability that Finland is one of them?

Answer 1

Solution

For convenience in presentation, let F, C, U, S, G and J represent Finland, Canada, USA, Switzerland, Germany and Japan respectively.

Back-up Theory

If a discrete random variable, X, has probability function, p(x), x = x₁, x₂, …., x_n, then

Mean (average) of X = E(X) = µ = Σ{x.p(x)} summed over all possible values of x…..…. (1)

E(X²) = Σ{x².p(x)} summed over all possible values of x…………………………………..(2)

Variance of X = Var(X) = σ² = E[{X – E(X)}²] = E(X²) – {E(X)}²………………....………..(3)

Standard Deviation of X = SD(X) = σ = sq.rt of Var(X) …………………..………………..(4)

Now, to work out the solution,

Game 1: Finland (0.2) vs. Canada (0.8)

Game 2: USA (0.3) vs. Switzerland (0.7)

Game 3: Germany (0.4) vs. Japan (0.6)
Part (a)

Sample space and probability

Under sample space, the letter represents the winners in the first, second and the third games in that order.

#	SS	Probability
1	FUG	0.2 x 0.3 x 0.4	= 0.024
2	FUJ	0.2 x 0.3 x 0.6	= 0.036
3	FSG	0.2 x 0.7 x 0.4	= 0.056
4	FSJ	0.2 x 0.7 x 0.6	= 0.084
5	CUG	0.8 x 0.3 x 0.4	= 0.096
6	CUJ	0.8 x 0.3 x 0.6	= 0.144
7	CSG	0.8 x 0.7 x 0.4	= 0.224
8	CSJ	0.8 x 0.7 x 0.6	= 0.336
		Total	1.000

Note: Sum of probabilities of any probability distribution MUST be one. CHECK

DONE Answer 1

Part (b)

European teams are: F, S and G

So, X can take 4 possible values, namely 0, 1, 2, 3

Probability Distribution X

x	p(x)	Row Referance of Above Table
0	0.144	6
1	0.468	2, 5, 8
2	0.332	1, 4, 7,
3	0.056	3
Total	1.000

Note: Sum of probabilities of any probability distribution MUST be one. CHECK

DONE Answer 2

Part (c)

Vide (1),

Expected value = 1.3 Answer 3

Vide (3),

Variance = 2.3 – 1.3²

= 0.61 Answer 4

Details

x	p(x)	Row Referance of Above Table	x.p(x)	x2.p(x)
0	0.144	6	0	0
1	0.468	2, 5, 8	0.468	0.468
2	0.332	1, 4, 7,	0.664	1.328
3	0.056	3	0.168	0.504
Total	1.000		1.300	2.3

Part (d)

Two European teams win => X = 2 [vide Part (b)]

Again, vide vide Part (b) Table,

Against, X = 2,the last column says 3 possibilities, namely,

1	FUG
4	FSJ
7	CSG

Out of 3 possibilities, F appears twice. Hence, the required probability = 2/3 Answer 5

DONE