# Thread: Probabilities and Odds Problems

1. ## Probabilities and Odds Problems

Hey,

Can someone help me on this plz and explain your solution.

A confident and boastful coach claims that on the next game the odds of her team winning are 3:1, the odds against losing are 5:1, and the odds against tieing are 7:1. Can these odds be right? Show why or why not these claims make sense.

Thanx

Ballin

2. Originally Posted by ballin_sensation
Hey,

Can someone help me on this plz and explain your solution.

A confident and boastful coach claims that on the next game the odds of her team winning are 3:1, the odds against losing are 5:1, and the odds against tieing are 7:1. Can these odds be right? Show why or why not these claims make sense.

Thanx

Ballin
I find it easiest to work these problems by rewriting the odds as probabilities.
The probability of winning is 3/4. Similarly, the probability of not losing is 5/6, and the probability of not tying is 7/8.
But the probability of tying should be the probability of not losing minus the probability of winning, which we find is:
$\displaystyle \frac{5}{6} - \frac{3}{4} = \frac{20}{24} - \frac{18}{24} = \frac{2}{24} = \frac{1}{12}$,
whence the probability of not tying is 11/12. So you would conclude that these odds do not make sense.

3. Thanx bro i get it. One more thing these probabilities are supposed to add up to 1 right?

4. Originally Posted by ballin_sensation
Thanx bro i get it. One more thing these probabilities are supposed to add up to 1 right?
Let P(W) represent the probability of winning;
let P(T) represent the probability of tying;
let P(L) represent the probability of losing.

Then $\displaystyle P(W) + P(T) + P(L) = 1$.

Note, however, that you are not explicitly given P(T) or P(L), though both P(T) and P(L) are easily computed from your given information.