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Math Help - Bayes Theorem

  1. #1
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    Bayes Theorem

    Hi there could anyone help me out with the following question please:

    A plane crash could be caused by 1 or 4 of the following events.

    H1 = faulty motor P(H1) = 0.2
    H2 = bomb in the plane P(H2) = 0.4
    H3 = missile that hit the plane P(H3) = 0.3
    H4 = fault in the computer system P(H4) = 0.1

    The investigation has shown that the crash was accompanied by the event A = fuel ignition. It is known that:
    P(A l H1 ) = 0.7;
    P(A l H2 ) = 0.9
    P(A l H3 ) = 0.8
    P(A l H4 ) = 0

    Calculate P(H2 l A )

    So i tried using the Bayes theorem but i got a probability of over 1. Could anyone help me out please.
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  2. #2
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    Quote Originally Posted by axa121 View Post
    Hi there could anyone help me out with the following question please:

    A plane crash could be caused by 1 or 4 of the following events.

    H1 = faulty motor P(H1) = 0.2
    H2 = bomb in the plane P(H2) = 0.4
    H3 = missile that hit the plane P(H3) = 0.3
    H4 = fault in the computer system P(H4) = 0.1

    The investigation has shown that the crash was accompanied by the event A = fuel ignition. It is known that:
    P(A l H1 ) = 0.7;
    P(A l H2 ) = 0.9
    P(A l H3 ) = 0.8
    P(A l H4 ) = 0

    Calculate P(H2 l A )

    So i tried using the Bayes theorem but i got a probability of over 1. Could anyone help me out please.
    Bayes' theorem for this says:

    p(H_2|A)=\frac{p(A|H_2)p(H_2)}{p(A)}

    p(A)=p(A|H_1)p(H_1)+p(A|H_2)p(H_2)+p(A|H_3)p(H_3)+  p(A|H_4)p(H_4)

    and if you do the arithmetic right p(H_2|A)<0.5.

    CB
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  3. #3
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    Hello, axa121!

    You left out a piece of data . . . I'll make one up.


    A plane crash could be caused by 1 or 4 of the following events:

    . . \begin{array}{cccccccc}<br />
\text{faulty motor} & P(\text{motor}) &=& 0.2 \\<br />
\text{bomb} & P(\text{bomb}) &=& 0.4 \\<br />
\text{missible} & P(\text{missile}) &=& 0.3 \\<br />
\text{computer} & P(\text{computer} &=& 0.1 \end{array}

    The crash was accompanied by the event \text{fuel ignition}.

    It is known that:

    . . \begin{array}{ccccc}<br />
P(\text{fuel }|\text{ motor}) &=& 0.7 & (1)\\<br />
P(\text{fuel }|\text{ bomb}) &=& 0.9 & (2)\\<br />
P(\text{fuel }|\text{ missile}) &=& 0.8 & (3)\\<br />
P(\text{fule }|\text{ computer}) &=& 0.6 & (4)\end{array}


    Calculate P(\text{bomb }|\text{ fuel})

    We want: . P(\text{bomb }|\text{ fuel}) \;=\; \frac{P(\text{bomb} \wedge \text{fuel})} {P(\text{fuel})} .(a)


    From (1), we have: . P(\text{fuel }|\text{ motor}) \:=\:\frac{P(\text{fuel}\wedge\text{motor})}{P(\te  xt{motor})} \:=\:0.7 \quad\Rightarrow\quad \frac{P(\text{fuel}\wedge\text{motor})}{.2} \:=\:0.7
    . . Hence: . P(\text{fuel}\wedge\text{motor}) \:=\:0.14

    From (2), we have: . P(\text{fuel }|\text{ bomb}) \;=\;\frac{P(\text{fuel}\wedge\text{bomb})}{P(\tex  t{bomb})}  \:=\:0.9\quad\Rightarrow\quad \frac{P(\text{fuel}\wedge\text{bomb})}{0.4} \:=\:0.9
    . . Hence: . P(\text{fuel}\wedge\text{bomb}) \:=\:0.36 .(b)

    From (3), we have: . P(\text{fuel }|\text{ missle}) \;=\;\frac{P(\text{fuel}\wedge\text{missile})}{P(\  text{missible})} \:=\:0.8 \quad\Rightarrow\quad \frac{P(\text{fuel}\wedge\text{missile})}{0.3} \:=\:0.8
    . . Hence: . P(\text{fuel}\wedge\text{missile}) \;=\;0.24

    Fromn (4), we have: . P(\text{fuel }|\text{ computer}) \:=\:\frac{P(\text{fuel}\wedge\text{computer})}{P(  \text{computer})} \:=\:0.6 \quad\Rightarrow\quad \frac{P(\text{fuel}\wedge\text{computer})}{0.1} \:=\:0.6
    . . Hence: . P(\text{fuel}\wedge\text{computer}) \:=\:0.06


    Then: . P(\text{fuel}) \:=\:0.14 + 0.36 + 0.24 + 0.06 \;=\;0.80 .p(c)


    Substitute (b) and (c) into (a): . P(\text{bomb }|\text{fuel}) \;=\;\frac{P(\text{bomb}\wedge\text{fuel})}{P(\tex  t{fuel})} \;=\;\frac{0.36}{0.80} \;=\;\boxed{0.45}

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  4. #4
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    Thanks, the way I had done it I got something like 67/35.

    The thing I am finding more difficlult with the Bayes Theorem is splitting the outcomes into partitions, but this question had already done so. I''l practice a few more examples.
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  5. #5
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    Soroban in the question it says P ( A l H4 ) = 0

    Thanks for the help though.
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