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Math Help - Probability help - 3 questions

  1. #1
    Newbie aj rules's Avatar
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    Probability help - 3 questions

    Hey there, hopefully someone is able to help me with these questions i am stuck with.
    I am in year 12 (final year)

    1. The chance that a harvest is poorer than average is 0.5, but if it is known that a certain disease D is present, this probability increases to 0.8. The disease D is present in 30% of the harvests. Find the probability that, when the harvest is observed to be poorer than average, the disease D is present.

    2. The test used to determine if a person suffers from a particular disease is not perfect. The probability of a person with the disease returning a positive result is 0.95, while the probability of a person without the disese returning a positive result is 0.02. The probability that a randomly selected person has the disease is 0.03. What is the probability that a randomly selected person will return a positive result?

    3. Nathan knows that his probability of kicking more than four goals on a wet day is 0.3, while on a dry day it is 0.6. The probability that it will be wet on the day of the next game is 0.7. Calculate the probability that Nathan will kick more than four goals in the next game.


    Thanks in advance.
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  2. #2
    MHF Contributor
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    Quote Originally Posted by aj rules View Post
    1. The chance that a harvest is poorer than average is 0.5
    What else would it be?

    Sorry, I was just a little busy laughing because I once hear the lead story on an Atlanta television news station say, during a drought, "The National Weather Service has predicted a 50% chance of above average rainfall in the next 30 days." I just laughed and laughed. (Okay, it's not the Median, but it's close enough for rainfall and for harvests.)

    Can you build a tree?

    Disease & Bad ==> .3*.8 = 0.24
    Disease & Good ==> .3*.2 = 0.06
    No Disease & Bad ==> .7*.5 = 0.35
    No Disease & Good ==> .7*.5 = 0.35

    "Find the probability that, when the harvest is observed to be poorer than average, the disease D is present."

    (Disease&Bad)/((Disease&Bad)+(NoDisease&Bad)) = 0.24/(0.24+0.35)
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  3. #3
    Newbie aj rules's Avatar
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    Quote Originally Posted by TKHunny View Post
    What else would it be?

    Sorry, I was just a little busy laughing because I once hear the lead story on an Atlanta television news station say, during a drought, "The National Weather Service has predicted a 50% chance of above average rainfall in the next 30 days." I just laughed and laughed. (Okay, it's not the Median, but it's close enough for rainfall and for harvests.)

    Can you build a tree?

    Disease & Bad ==> .3*.8 = 0.24
    Disease & Good ==> .3*.2 = 0.06
    No Disease & Bad ==> .7*.5 = 0.35
    No Disease & Good ==> .7*.5 = 0.35

    "Find the probability that, when the harvest is observed to be poorer than average, the disease D is present."

    (Disease&Bad)/((Disease&Bad)+(NoDisease&Bad)) = 0.24/(0.24+0.35)
    Genious! haha thanks for the help.
    How did you know the end forumla? Has it got something to do with Pr(AUB) and all that? Also, could a ven diagram have been made?
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  4. #4
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    Quote Originally Posted by aj rules View Post
    Genious! haha thanks for the help.
    How did you know the end forumla? Has it got something to do with Pr(AUB) and all that? Also, could a ven diagram have been made?
    Drawing an appropriate tree diagram (for all three questions) is the best approach since tree diagrams have built into them the conditional probability data you've been given. A Venn diagram is a clumsy approach and is NOT prefered for this type of given data (a Karnaugh Table would be slightly less clumsy).
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