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Math Help - Probability that weather is fine

  1. #1
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    Probability that weather is fine

    In a simple model of the weather in october, each day is classified as either fine or rainy. The oribability that a fine day is followed by a fine day is 0.8. The probability that a rany day is followed by a fine day is 0.4. The probability that 1 October is fine is 0.75.

    Find the probability that 2 october is fine and the probability that 3 october is fine.

    P(2oct is fine and 3oct is fine)=0.75(0.8)(0.8)=0.48

    but answer states 0.56

    Also, I need help with: Find the conditional probability that 3 October is rainy, given that 1 October is fine.
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    Re: Probability that weather is fine

    You have F for fine and R for rainy, then

    F_{n+1} =0.8 F_{n}+0.4 R_{n}
    R_{n+1} =0.2F_{n}+0.6R_{n}

    Now F_{n}=0.75, R_{n}= 0.25

    Iterate twice
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  3. #3
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    Re: Probability that weather is fine

    Quote Originally Posted by pickslides View Post
    You have F for fine and R for rainy, then

    F_{n+1} =0.8 F_{n}+0.4 R_{n}
    R_{n+1} =0.2F_{n}+0.6R_{n}

    Now F_{n}=0.75, R_{n}= 0.25

    Iterate twice
    Sorry, but I don't fully understand your solution.

    Am I right to say F (n+1) means the probability that oct 2 is a fine day? What about october 3 then?
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    Re: Probability that weather is fine

    Quote Originally Posted by Punch View Post
    Sorry, but I don't fully understand your solution.

    Am I right to say F (n+1) means the probability that oct 2 is a fine day? What about october 3 then?
    Yep, thats the way!

    Use the results from oct 2 to find for oct 3. same formula.
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    Re: Probability that weather is fine

    Quote Originally Posted by pickslides View Post
    Yep, thats the way!

    Use the results from oct 2 to find for oct 3. same formula.
    You stated F(n+1) = 0.8F(n)+0.4R(n)

    Why is it that R(n) has to be included? Why is it not F(n+1)=0.8F(n)
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    Re: Probability that weather is fine

    Quote Originally Posted by Punch View Post
    You stated F(n+1) = 0.8F(n)+0.4R(n)

    Why is it that R(n) has to be included? Why is it not F(n+1)=0.8F(n)
    Because a fine day can preceed both a fine day or a rainy day.
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    Re: Probability that weather is fine

    Quote Originally Posted by pickslides View Post
    Because a fine day can preceed both a fine day or a rainy day.
    I see... I can't believe I overlooked that!

    P(Fine on oct 2 & oct 3)= P(FFF)+P(RFF) =0.75(0.8)(0.8)+0.25(0.4)(0.8)=0.56

    But there's still a second part I have a problem with in post 1
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    Re: Probability that weather is fine

    Have you looked at P(A/B) = \frac{P(A/cap B)}{P(B)} ?
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    Re: Probability that weather is fine

    Its the exact same forumla pickslides gave you. Except instead of oct 1 being a 75/25 split (Fn=.75 and Rn=.25) you know its clear. So Fn=1.0 and Rn =0.0
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  10. #10
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    Lightbulb Re: Probability that weather is fine

    The original question is quoted from "Advanced Level Statistics - J. Crawshaw & J. Chambers, Exercise 3d Tree diagram, Section B, Question 10 (Page 202)."
    I had to get through a lot of ambiguous solutions before finally get the answers perfectly matched with those of the textbook.

    In a simple model of the weather in October, each day is classified as either fine or rainy. The probability that a fine day is followed by a fine day is 0.8. The probability that a rainy day is followed by a fine day is 0.4 The probability that 1 October is fine is 0.75.

    (a) Find the probability that 2 October is fine and the probability that 3 October is fine. answer: 0.7, 0.68
    (b) Fine the conditional probability that 3 October is rainy, given that 1 October is fine. answer: 0.28
    (c) Find the conditional probability that 1 October is fine, given that 3 October is rainy. answer: 0.65625
    (C)

    Probability that weather is fine-my-tree-diagram.jpg

    (a)
    (i) P(F_{1}\cap F_{2})+P(R_{1}\cap F_{2})
    = (0.75\times 0.8) + (0.25\times 0.4) =0.7

    (ii)
    P(F_{1}\cap F_{2}\cap F_{3})+P(F_{1}\cap R_{2}\cap F_{3})+P(R_{1}\cap F_{2}\cap F_{3})+P(R_{1}\cap R_{2}\cap F_{3})
    = (0.75\times 0.8\times0.8) + (0.75\times0.2\times0.4) + (0.25\times0.4\times0.8) + (0.25\times0.6\times0.4) =0.68


    (b)
    \frac{P(F_{1}\cap F_{2}\cap R_{3})+P(F_{1}\cap R_{2}\cap R_{3})}{P(F_{1}\cap F_{2}\cap F_{3})+P(F_{1}\cap F_{2}\cap R_{3})+P(F_{1}\cap R_{2}\cap F_{3})+P(F_{1}\cap R_{2}\cap R_{3})}

    = \frac{(0.75\times 0.8\times0.2) + (0.75\times0.2\times0.6)}{(0.75\times0.8\times0.8) + (0.75\times0.8\times0.2) + (0.75\times0.2\times0.4) + (0.75\times0.2\times0.6)} =0.28


    (c) \frac{P(F_{1}\cap F_{2}\cap R_{3})+P(F_{1}\cap R_{2}\cap R_{3})}{P(F_{1}\cap F_{2}\cap R_{3})+P(F_{1}\cap F_{2}\cap R_{3})+P(R_{1}\cap F_{2}\cap R_{3})+P(R_{1}\cap R_{2}\cap R_{3})}

    = \frac{(0.75\times 0.8\times0.2) + (0.75\times0.2\times0.6)}{(0.75\times0.8\times0.2) + (0.75\times0.2\times0.6) + (0.25\times0.4\times0.2) + (0.25\times0.6\times0.6)} = 0.65625
    Last edited by zikcau25; January 21st 2014 at 11:07 AM.
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