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Math Help - Conditional probability

  1. #1
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    Conditional probability

    Suppose we have 2 events: s, e.

    We know:

    P(e)=.7
    P(e|\neg s)=.15
    P(e|s)=1

    What is P(s)?

    ----------

    I know that P(s) = 1 - P(\neg s) \Rightarrow P(\neg s)=1-P(s) and that

    P(e|s) = \frac{P(s|e)P(e)}{P(s)}=\frac{P(s|e)P(e)}{1-P(\neg s)}=\frac{.7P(s|e)}{1-P(\neg s)}=1

    \Rightarrow .7P(s|e)=1-P(\neg s)

    \Rightarrow P(\neg s)=1-.7P(s|e)

    \Rightarrow P(s) = .7P(s|e)

    but I sort of go around in circles...
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  2. #2
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    Quote Originally Posted by scorpion007 View Post
    Suppose we have 2 events: s, e.

    We know:

    P(e)=.7
    P(e|\neg s)=.15
    P(e|s)=1

    What is P(s)?

    ----------

    I know that P(s) = 1 - P(\neg s) \Rightarrow P(\neg s)=1-P(s) and that

    P(e|s) = \frac{P(s|e)P(e)}{P(s)}=\frac{P(s|e)P(e)}{1-P(\neg s)}=\frac{.7P(s|e)}{1-P(\neg s)}=1

    \Rightarrow .7P(s|e)=1-P(\neg s)

    \Rightarrow P(\neg s)=1-.7P(s|e)

    \Rightarrow P(s) = .7P(s|e)

    but I sort of go around in circles...
    hi

    P(e|s)=P(S n e)/p(s)

    P(S n e)= P(s) ---1

    P(e|s')=P(s' n e)/P(s')

    0.15=[P(e)-P(S n e)]/[1-P(s)]

    0.15[1-P(s)]=0.7-P(s) from 1
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  3. #3
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    Hello, scorpion007!

    \text{Suppose we have two events: }\:e,\, s.

    We know: . \begin{Bmatrix}P(e) &=& 0.7 & [1]\\<br />
P(e\,|\sim\!s) &=& 0.15 & [2]\\<br />
P(e \,|\, s) &=& 1 & [3] \end{Bmatrix}

    What is P(s) ?

    I filled in a chart . . .


    From [1]: . P(e) \:=\:0.7 \quad\Rightarrow\quad P(\sim\!e) \:=\:0.3

    . . \begin{array}{c||c|c||c|}<br />
& P(s) & P(\sim\!s) & \text{Total} \\ \hline\hline<br />
P(e) & & & {\color{red}0.7} \\ \hline<br />
P(\sim\!e) & & & {\color{red}0.3} \\ \hline\hline<br />
\text{Total} & {\color{red}P(s)} & {\color{red}P(\sim\!s)} & 1.0 \end{array}



    From [3]: . P(e\,|\,s) \:=\:\frac{P(e\,\wedge\,s)}{P(s)} \:=\:1

    Hence: . P(e\wedge s) \:=\:P(s) \quad\Rightarrow\quad P(\sim\!e\wedge s) \:=\:0

    . . \begin{array}{c||c|c||c|}<br />
& P(s) & P(\sim\!s) & \text{Total} \\ \hline\hline<br />
P(e) & {\color{red}P(s)}  & & 0.7 \\ \hline<br />
P(\sim\!e) & {\color{red}0} & & 0.3 \\ \hline\hline<br />
\text{Total} & P(s) & P(\sim\!s) & 1.0 \end{array}



    From [2]: . P(e\,|\sim\!s) \:=\:\frac{P(e \wedge\sim\!s)}{P(\sim\!s)} \:=\:0.15

    Hence: . P(e\,\wedge \sim\!s) \:=\:0.15P(\sim\!s) \quad\Rightarrow\quad P(\sim\!e \wedge \sim\!s) \:=\:0.85P(\sim\!s)

    . . \begin{array}{c||c|c||c|}<br />
& P(s) & P(\sim\!s) & \text{Total} \\ \hline\hline<br />
P(e) & P(s) & {\color{red}0.15P(\sim\!s)} & 0.7 \\ \hline<br />
P(\sim\!e) & 0 & {\color{red}0.85P(\sim\!s)} & 0.3 \\ \hline\hline<br />
\text{Total} & P(s) & P(\sim\!s) & 1.0 \end{array}


    Reading across, we have: . \begin{array}{ccccc}<br />
P(s) + 0.15P(\sim\!s) &=& 0.7 & [1] \\<br />
0 + 0.85P(\sim\!s) &=& 0.3 & [2] \end{array}

    From [2], we have: . 0.85P(\sim\!s) \:=\:0.3 \quad\Rightarrow\quad P(\sim\!s) \:=\:\frac{0.3}{0.85} \:=\:\frac{6}{17}


    Therefore: . P(s) \;=\;1-\frac{6}{17} \;=\;\frac{11}{17}

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  4. #4
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    thank you both very much!
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  5. #5
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    Quote Originally Posted by mathaddict View Post
    0.15=[P(e)-P(S n e)]/[1-P(s)]

    is it true in general that

    P(A\cap B) = P(A) - P(A\cap \neg B) ?
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  6. #6
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    Quote Originally Posted by scorpion007 View Post
    is it true in general that

    P(A\cap B) = P(A) - P(A\cap \neg B) ?
    in general P(A n B')=P(A)-P(A n B)

    It should be clear when you draw the sets
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