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Math Help - Probability question related to the tree diagram

  1. #1
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    Probability question related to the tree diagram

    "Mike races with another cyclist, the probability that Mike wins is 0.8 when it is a rainy day, and the probability that Mike wins on a dry day is 0.3. The probability that it is a rainy day is 0.3
    -Find the probability that Mike wins the race"

    From the example, it drew a tree diagram and showed that the answer would be 0.3 x 0.8 + 0.7 x 0.3 = 0.45.

    But from my logical point of view, this should also give the right answer: P(M|R) + P(M|D), where the first is the probability of him wining when its raining and the probability of him wining when its given that its not rainy (dry).

    So, it should be: Intersection of M and R divided by P(R) + Intersection of M and D divided by P(D), and it is given that the intersection when it rains is 0.8 and of when its dry is 0.3, so I dont understand how this would be wrong: 0.8/0.3 + 0.3/0.7.


    Regards,
    Zananok
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  2. #2
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    Re: Probability question related to the tree diagram

    Quote Originally Posted by Zananok View Post
    "Mike races with another cyclist, the probability that Mike wins is 0.8 when it is a rainy day, and the probability that Mike wins on a dry day is 0.3. The probability that it is a rainy day is 0.3
    -Find the probability that Mike wins the race"
    From the example, it drew a tree diagram and showed that the answer would be 0.3 x 0.8 + 0.7 x 0.3 = 0.45.
    But from my logical point of view, this should also give the right answer: P(M|R) + P(M|D), where the first is the probability of him wining when its raining and the probability of him wining when its given that its not rainy (dry).
    So, it should be: Intersection of M and R divided by P(R) + Intersection of M and D divided by P(D), and it is given that the intersection when it rains is 0.8 and of when its dry is 0.3, so I dont understand how this would be wrong: 0.8/0.3 + 0.3/0.7.
    You are using the formula incorrectly.
    \mathcal{P}(M)=\mathcal{P}(M\cap R)+\mathcal{P}(M\cap D)=\mathcal{P}(M|R)\mathcal{P}(R)+\mathcal{P}(M|D)  \mathcal{P}(D)=(0.8)(0.3)+(0.2)(0.7)
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  3. #3
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    Re: Probability question related to the tree diagram

    Thank you, that explains everything.


    Regards,
    Thomas Figueiredo Jøssund
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