Probability question

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March 9th 2010, 04:48 AM
paulig

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...
March 9th 2010, 01:32 PM
Soroban

Hello, paulig!

This is long, tedious problem . . .

Quote:

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.

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

Then add these 16 decimals.

I'll wait in the car . . .
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March 10th 2010, 02:13 AM
paulig

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
March 10th 2010, 01:45 PM
awkward

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.