Thread: 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...

Hello, paulig!

This is long, tedious problem . . .

There are 5 objects that could be "on".

The probabilities for each being "on" is: .

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.

. .


Four are on: there are 5 cases.

. .


Three are on: there are 10 cases.

. .


Then add these 16 decimals.

I'll wait in the car . . .
.

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

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.