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Math Help - Odds

  1. #1
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    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.
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by BreedingtheSpawn View Post
    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 \alpha, of an event, \displaystyle A, is defined by:

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

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

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

    \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?
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  3. #3
    Member Veronica1999's Avatar
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    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.
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