# Thread: Odds

1. ## Odds

In a certain town, 10% of people commute to work by bicycle. If a person is selected randomly from the town, what are the odds in favor of selecting someone who commutes by bicycle?

1 to 9
9 to 1
1 to 10
10 to 9

Any help on this would be appreciated.

2. Originally Posted by BreedingtheSpawn
In a certain town, 10% of people commute to work by bicycle. If a person is selected randomly from the town, what are the odds in favor of selecting someone who commutes by bicycle?

1 to 9
9 to 1
1 to 10
10 to 9

Any help on this would be appreciated.
The odds, $\displaystyle \displaystyle \alpha$, of an event, $\displaystyle \displaystyle A$, is defined by:

$\displaystyle \displaystyle \alpha = \frac {P(A)}{1 - P(A)}$

where $\displaystyle \displaystyle P(A)$ is the probability of $\displaystyle \displaystyle A$ occuring, and we say the odds are $\displaystyle \displaystyle \alpha :1$ in favor of the event.

For example, lets say you knew the probabilty for some event $\displaystyle \displaystyle A$ occuring is $\displaystyle \displaystyle P(A) = \frac 13$. What are the odds in favor of such an event? Since

$\displaystyle \displaystyle \alpha = \frac {\frac 13}{1 - \frac 13} = \frac 12$

The odds in favor are 1/2 : 1, or to clear fractions, multiply both sides by 2 and say the odds in favor for A are 1 : 2 (it's more likely for A not to occur, twice as likely for it to not occur than occur)

Now, apply this to solving your problem. What can you come up with?

3. The odds favoring an event occuring is the ratio of the probability that it does happen to the probability that it does not happen.
Let's say the town has a 100 people then the ratio of the people who commute by bicycle to the people who do not is 10 /100: 90/100 which is basically 1:9. ( Odds almost never contain fractions)

If the problem asked for the odds in favor of selecting someone who does not commute by bicycle the answer would be 9:1.