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Math Help - Probability question

  1. #1
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    Probability question

    Hey, lets say I have 5 probabilities of the chance that 5 different things are is "on", so the probabilities for each of the items would be something like:

    .75, .81, .63, .9, .1

    I'd like to calculate the probability that, given the individual probabilities, that 3 or more of these items are really "on".

    I'm guessing the probability that all 5 are "on" is simply the 5 numbers multiplied together, which is ~.03.

    I'd imagine I'd have to add this number to the probability that 4 are and on the probability that 3 of them are on to get the total probability for all possible outcomes? Not sure how to calculate these though...
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  2. #2
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    Hello, paulig!

    This is long, tedious problem . . .


    There are 5 objects \{A,B,C,D,E\} that could be "on".

    The probabilities for each being "on" is: . \begin{array}{ccc} P(A) &=& 0.75 \\ P(B) &=& 0.81 \\ P(C) &=& 0.63 \\ P(D) &=& 0.90 \\ P(E) &=& 0.10 \end{array}

    Find the probability that 3 or more of these items are "on".
    Your game plan is correct!


    All five are on: there is one case.

    . . P(A \cap B \cap C \cap D \cap E) \;=\;(0.75)(0.82)(0.63)(0.90)(0.10) \;=\;0.3444525


    Four are on: there are 5 cases.

    . . \begin{array}{ccccc} P(A \cap B \cap C \cap D) &=& (0.75)(0.81)(0.63)(0,90) &=& 0.334 452 5 \\ P(A \cap B \cap D \cap E) &=& (0.75)(0.81)(0.63)(0.10 &=& 0.038 2725 \\ \vdots && \vdots && \vdots \\ P(B \cap C \cap D \cap E) &=& (0.81)(0.63)(0.90)(0.10) &=& 0.045 9270  \end{array}


    Three are on: there are 10 cases.

    . . \begin{array}{ccccc}<br />
P(A \cap B \cap C) &=& (0.75)(0.81)(0.63) &=& 0.382725 \\<br />
P(A \cap B \cap D) &=& (0.75)(0.81_)0.90) &=& 0.546750 \\<br />
P(A \cap B \cap E) &=& (0.75)(0.81)(0.10) &=& 0.060750 \\<br />
P(A \cap C \cap D) &=& (0.75)(0.63)(0.90) &=& 0.425250 \\<br />
\vdots && \vdots && \vdots \\<br />
P(C \cap D \cap E) &=& (0.63)(0.90)(0.10) &=& 0.056700 \end{array}


    Then add these 16 decimals.


    I'll wait in the car . . .
    .
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  3. #3
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    Thanks for your reply. Ok so that's it then, I guess I posted it cause I thought there might be an easier way to calculate this, some kind of mathematical tool to do the job easily, unfortunately that doesn't seem to be the case, thanks again!

    P
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  4. #4
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    Just a comment on Soroban's solution--

    This solution assumes the events are independent. There is nothing in the problem statement to this effect. If it is *not* the case that the events are independent, then you do not have sufficient information to solve the problem.
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