3 athletes running

Question

Three athletes $A_1, A_2, A_3$ run a race.

Let $E_{ij}$ denotes the event that athlete $A_i$ arrives before athlete $A_j$, and let $E_{ijk}$ denotes the event that athlete $A_i$ arrives before athlete $A_j$ who arrives before athlete $A_k$.

Given that \begin{gather} P(E_{13})=2/3;\\P(E_{23})=1/2;\\P(E_{123})=P(E_{132})=x;\\P(E_{213})=P(E_{231})=y;\\ P(E_{312})=P(E_{321})=z.\end{gather}

1. Find $x, y,z$.
2. Calculate the probability that athlete A₁ comes first.
3. Are the events $E_{13}$ and $E_{23}$ independent ?

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Any tips to solve this are greatly appreciated!!
What I understand :
3. I could show that $P(E_{13}\cap E_{23})= P(E_{13}) P(E_{23})$

What I don't understand:

1. What does $P_{123}$ mean?
2. Is $P(E_1\cap E_3) \neq\varnothing$ ?

Answer 1

I've figured it out eventually:
(a) We know that: E₁₃ = E₁₂₃ $\cup$ E₂₁₃ $\cup$ E₁₃₂
Thus: P(E₁₂₃)+ P(E₂₁₃) + P(E₁₃₂) = $2\over3$ $\implies$ 2x + y = $2\over3$ (1)
also: E₂₃ = E₁₂₃ $\cup$ E₂₃₁ $\cup$ E₂₁₃ and:
P(E₁₂₃) + P(E₂₁₃) + P(E₂₃₁) = $1\over2$ $\implies$ 2y + x = $1\over2$ (2).
Solving them system of (1),(2) we get:
y = $1\over9$ and x = $5\over18$ .
However (all the possible outcomes are 3! = 6) :
P( $\bigcup$ E _ijk) = 1 $\implies$ x + x + y + y + z + z =1
$\implies$ x + y + z = $1\over2$ (3)
and by substituting the values of x,y to (3) we get: z = $11\over18$.
(b) P( E₁₂₃ $\cup$ E₁₃₂) = P(E₁₂₃) + P(E₁₃₂) = 2x = $5\over 18$ = $5\over 9$
(c) P(E₁₃) P(E₂₃) = $2\over3$ $1\over2$ = $1\over 3$ (4) and:
E₁₃ = {E₁₂₃ , E₁₃₂, E₂₁₃ }
E₂₃ = {E₁₂₃, E₂₁₃, E₂₃₁ }. These sets have in common 2 events so:
P(E₁₃ $\cap$ E₂₃)= $2\over 6$ = $1\over3$ (5)
(4),(5) $\implies$ P(E₁₃ $\cap$ E₂₃)= P(E₁₃) P(E₂₃), which means that events E₁₃ and E₂₃ are independent.

Answer 2

You can see that the situation is interesting: The first two formulas seem to indicate that A1 is the best runner, while A2 and A3 Are equally good. But the last three say that whoever wins, the other two are equally likely to beat each other. For example, A1 usually beats A3, but not if A2 wins the race.

How can this happen? Say A1 is good in sunshine, A2 is excellent in bad weather, A3 suffers in bad weather, but not as much as A1. So if A2 wins, the weather is likely bad, and A3s chances to beat A1 are better than usual.

Find P(E31) and P(E32), which is very, very easy, set up whatever equations you find for x, y and z, and try to solve them.