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Math Help - probability - plz help

  1. #1
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    Smile probability - plz help

    1. If P(T) = P(Q) =3(T intersection Q) and P(T U Q) = 0.75
    Find
    (a) P(T intersection Q)
    (b) P(T)
    (c) P(Q')

    2. Two dice are thrown. One die lands on 2. Find the probability that product of dice is
    (a) exactly 6
    (b) more than 5

    Any help or hint much appreciated

    Thanks in advance

    Dee
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  2. #2
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    Quote Originally Posted by dee2020 View Post
    1. If P(T) = P(Q) =3(T intersection Q) and P(T U Q) = 0.75
    Find
    (a) P(T intersection Q)
    (b) P(T)
    (c) P(Q')

    2. Two dice are thrown. One die lands on 2. Find the probability that product of dice is
    (a) exactly 6
    (b) more than 5
    1) P(T\cup Q)=P(T)+P(Q)-P(T\cap Q)

    2) There are thirty-six possible pairs.
    The event where the product equals six is \{(2,3),(3,2)\}
    SO?
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  3. #3
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    hi, thanks for your reply
    for 2nd question, do i have to use conditional probability formula
    P(B|A) = P(B intersection A)/ P(A)
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  4. #4
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    Quote Originally Posted by dee2020 View Post
    hi, thanks for your reply
    for 2nd question, do i have to use conditional probability formula
    P(B|A) = P(B intersection A)/ P(A)
    Eleven pairs contain a two. So \frac{2}{11}
    Last edited by Plato; April 26th 2010 at 03:34 PM.
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  5. #5
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    Quote Originally Posted by Plato View Post
    Eleven pairs contain a two. So \frac{2}{11}
    Thx very much again, could you please briefly explain how did you get 2/11, how did you work out P(B intersection A) in the formula?
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  6. #6
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    Quote Originally Posted by dee2020 View Post
    Thx very much again, could you please briefly explain how did you get 2/11, how did you work out P(B intersection A) in the formula?
    Here are the thirty-six outcomes:
    (1,1), (1,2), (1,3), (1,4), (1,5), (1,6)
    (2,1), (2,2), (2,3), (2,4), (2,5), (2,6)
    (3,1), (3,2), (3,3), (3,4), (3,5), (3,6)
    (4,1), (4,2), (4,3), (4,4), (4,5), (4,6)
    (5,1), (5,2), (5,3), (5,4), (5,5), (5,6)
    (6,1), (6,2), (6,3), (6,4), (6,5), (6,6)
    You are told that there is a two in the pair.
    So there are only eleven pairs contain a two.
    Of those only two pairs multiply to six.
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  7. #7
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    Quote Originally Posted by Plato View Post
    Here are the thirty-six outcomes:
    (1,1), (1,2), (1,3), (1,4), (1,5), (1,6)
    (2,1), (2,2), (2,3), (2,4), (2,5), (2,6)
    (3,1), (3,2), (3,3), (3,4), (3,5), (3,6)
    (4,1), (4,2), (4,3), (4,4), (4,5), (4,6)
    (5,1), (5,2), (5,3), (5,4), (5,5), (5,6)
    (6,1), (6,2), (6,3), (6,4), (6,5), (6,6)
    You are told that there is a two in the pair.
    So there are only eleven pairs contain a two.
    Of those only two pairs multiply to six.
    thx u soooooo much..
    much appreciated :-)
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