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Thread: Help needed for a probablilty question

  1. #1
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    Angry Help needed for a probablilty question

    Ok so this is really bugging me, because I thought i finally understood it, but then my book is giving me different answers to what I'm getting.
    Anyways, heres the problem, can anyone answer it and show working , so I can learn. Thanks

    Given P(R|S) = 0.5, P(R|S') = 0.4 and P(S) = 0.6 find:

    (a) P(R) (b) P(S|R) (c) P(S'|R) (d) P(S'|R')
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  2. #2
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    P(R)
    = P(R(S+S'))
    = P(R1)
    = P(RS)+P(RS')
    = P(R|S)P(S)+P(R|S')P(S')
    = P(R|S)P(S)+P(R|S')(1-P(S))

    P(RS)
    = P(R|S)P(S)
    = P(S|R)P(R)

    Now you know P(R) and P(RS), so you know P(RS')
    P(RS')
    = P(S'|R)P(R)

    P(S)
    = ...
    = P(S|R)P(R)+P(S|R')(1-P(R))
    gives
    P(SR')
    = P(S|R')P(R')
    = P(S|R')(1-P(R))
    Using this,
    P(1)
    = P((R+R')(S+S'))
    = P(RS) + P(RS')+P(R'S)+P(R'S')
    gives
    P(R'S')
    = P(S'|R')P(R')
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  3. #3
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    Quote Originally Posted by kamicool17 View Post
    Given P(R|S) = 0.5, P(R|S') = 0.4 and P(S) = 0.6 find:
    (a) P(R) (b) P(S|R) (c) P(S'|R) (d) P(S'|R')
    $\displaystyle P(RS)=P(R|S)P(S)$
    $\displaystyle P(RS')=P(R|S')P(S')$
    $\displaystyle P(R)=P(RS)+P(RS')$

    Now finish.
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  4. #4
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    Hello, kamicool17!

    Another approach . . . (much longer, though)


    Given: .$\displaystyle P(R|S) = 0.5,\;\;P(R|S') = 0.4,\;\;P(S) = 0.6$

    Find: . $\displaystyle (a)\;P(R) \qquad (b)\;P(S|R)\qquad (c)\;P(S'|R) \qquad (d)\;P(S'|R')$

    We have this table:

    . . $\displaystyle \begin{array}{c|| c|c||c}
    & S & S' & \text{Total } \\ \hline \hline
    R & {\color{blue}(a)} & {\color{blue}(c)} & {\color{blue}(e)} \\ \hline
    R' & {\color{blue}(b)} & {\color{blue}(d)} & {\color{blue}(f)} \\ \hline \hline
    \text{Total} & 0.60 & 0.40 & 1.00 \end{array}$

    We have: .$\displaystyle P(S) = 0.60 \quad\Rightarrow\quad P(S') = 0.40$
    So we can fill in the bottom row.



    We have: .$\displaystyle P(R|S) = 0.5$

    Then: .$\displaystyle \frac{P(R \wedge S)}{P(S)} \;=\;0.5 \quad\Rightarrow\quad \frac{P(R \wedge S)}{0.6} \:=\:0.5
    $

    Hence: .$\displaystyle P(R \wedge S) \,=\,0.30\;{\color{blue}(a)} \quad\Rightarrow\quad P(R' \wedge S) \:=\:0.30\;{\color{blue}(b)} $

    . . $\displaystyle \begin{array}{c|| c|c||c}
    & S & S' & \text{Total } \\ \hline \hline
    R & 0.30 & (c) & (e) \\ \hline
    R' & 0.30 & (d) & (f) \\ \hline \hline
    \text{Total} & 0.60 & 0.40 & 1.00 \end{array}$



    We have: .$\displaystyle P(R|S') = 0.4$

    Then: .$\displaystyle \frac{P(R \wedge S')}{P(S')} \:=\:0.4 \quad\Rightarrow\quad \frac{P(R \wedge S')}{0.4} \:=\:0.4$

    Hence: .$\displaystyle P(R \wedge S') \:=\:0.16\;{\color{blue}(c)} \quad\Rightarrow\quad P(R' \wedge S') \:=\:0.24\;{\color{blue}(d)} $

    Also: .$\displaystyle P(R) \:=\:0.46 \;{\color{blue}(e)}\quad\Rightarrow\quad P(R') \:=\:0.54\;{\color{blue}(f)}$



    And we have completed the table:

    . . $\displaystyle \begin{array}{c|| c|c||c}
    & S & S' & \text{Total } \\ \hline \hline
    R & 0.30 & 0.16 & 0.46 \\ \hline
    R' & 0.30 & 0.24 & 0.54 \\ \hline \hline
    \text{Total} & 0.60 & 0.40 & 1.00 \end{array}$

    You can now answer the questions . . .

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