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Math Help - A few basic probability question

  1. #1
    Senior Member Danneedshelp's Avatar
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    A few basic probability question

    Q: A large group of people is to be checked for two common symptoms of a certain disease. It is thought that 20% of the people possess symptom A alone, 30% possess symptom B alone, 10% possess both symptoms, and the remainder have neither symptom. For one person chosen at random from this group, find these probabilitys:

    a) The person has neighter symptom.

    b) The person has at least one symptom.

    c) The person has both symptoms, given that he has symptom B.

    A: I am assuming these events are independent.

    a) P(A\cup\\B)^{c}=1-P(A\cup\\B)=1-[P(A)+P(B)-P(A\cap\\B)]=<br />
1-[.2+.3-.1]=1-.4=.6

    b) P(A\cup\\B)=P(A)+P(B)-P(A\cap\\B)=.2+.3-.1=.4

    c) P(A\cap\\B|B)=\frac{P[(A\cap\\B)\cap\\B]}{P(B)}<br />
=\frac{P[(A\cap\\B)\cap\\(B\cap\\B)]}{P(B)}<br />
=\frac{P(A\cap\\B)P(B)}{P(B)}=.33333

    The book says teh answers are:

    a).4
    b).6
    c).25

    I am not sure what I am doing wrong.

    Thanks
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  2. #2
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    Quote Originally Posted by Danneedshelp View Post
    Q: A large group of people is to be checked for two common symptoms of a certain disease. It is thought that 20% of the people possess symptom A alone, 30% possess symptom B alone, 10% possess both symptoms, and the remainder have neither symptom. For one person chosen at random from this group, find these probabilitys:
    You have missread the question.
    The point is alone: P(A\cap B^c)=0.2,~P(B\cap A^c)=0.3~\&~P(A\cap B)=0.1.

    So P(A)=0.3~\&~P(B)=0.4.

    Redo you caculations.
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