A coach claims that on the next league game the odds of her team winning are 3:1; the odds against losing are 5:1; and the odds against a tie are 7:1. Can these odds be right? Explain.

Answer 1

odds = probability yes / probability no
so
prob win/prob lose = 3/1
however we know that prob not win (lose or tie)= 1 - prob win
so
w /(1-w) = 3
w = 3 - 3w
w =3/4 = 0.75 = probabiity of win = P(W)
0.25 = prob lose or tie = P(L+T)

odds win or tie = 5/1
prob win or tie / prob lose = 5
P(L) = [P(W)+P(T)] / 5 = [.75 +P(T)] / 5

odds no tie = odds win or lose = 7/1
prob win or lose / prob tie = 7
P(T) = [P(W)+P(L)] / 7 = [ .75 + P(L)] /7

now P( W) + P(L) + P(T) must = 1
.75 + [.75 +P(T)] / 5 + [ .75 + P(L)] /7 = 1?
.75 ( 1 + .2 + .142) + .2 P(T) + .142 P(L) = 1 ?
1.0065 + .2 P(T) + .142 P(L) = 1 ?
Nope, not unless one of those probabilities is negative