SOLUTION: A machine has fifteen identical components that function independently. It will stop working if three or more of the components fail. If the probability that an individual componen

Question 1038464: A machine has fifteen identical components that function independently. It will stop working if three or more of the components fail. If the probability that an individual component fails is 0.08, find the probability that the machine will be working

Found 3 solutions by Boreal, solver91311, Theo:
Answer by Boreal(15235)   (Show Source): You can put this solution on YOUR website!
Find probability that 0,1,2 don't work.
(0.92^15)=0.2863
15*(0.92^14*0.08=0.3734
15C2(0.92^13))0.08^2)=0.2273
Add the 3: 0.887 probability the machine will be working.

Answer by solver91311(24713)   (Show Source): You can put this solution on YOUR website!

You want the probability that the machine will be working, so you want the probability of 2 or fewer occurrences of a failure in 15 trials given the probability of success on any given trial is 0.08.

The probability of successes in trials where is the probability of success on any given trial is given by:

Where is the number of combinations of things taken at a time and is calculated by

Since you are looking for the probability of 2 or fewer, you need the probability of 0 plus the probability of 1 plus the probability of 2, that is:

An easier way to get the answer, given that you have MS Excel on your PC or Numbers on your MAC, is to open a spreadsheet and type the following:

=BINOMDIST(2,15,0.08,TRUE)

John

My calculator said it, I believe it, that settles it

Answer by Theo(13342)   (Show Source): You can put this solution on YOUR website!
if 3 or more of the components fail, then it will stop working.
that means it will continue working of 0, 1, or 2 of the components fail.
the probability that an individual component fails is .08.
the probability that an individual component does not fail is .92.
this is a binomial probability event with p = .08 and q = .92
p(x) = c(n,x) * p^x * q^(n-x)
x is the number of components that will fail.
p is the probability that an individual component will fail.
q is the probability that an individual component will not fail.
n is the total number of components.
p(x) is the probability that x components will fail.

the probability that 0 components will fail is p(0) = c(15,0) * .08^0 * .92^15.
the probability that 1 component will fail is p(1) = c(15,1) * .08^1 * .92^14.
the probability that 2 components will fail is p(2) = c(15,2) * .08^2 * .92^13.

you get:
p(0) = .286297...
p(1) = .373431...
p(2) = .227306...
p(0 + 1 + 2) = .887034...
this can be rounded off to .8870 which is equal to 88.70%

this means that the probability of the machine working at any given point in time is 88.70% which means that, in the long run, is should be working 88.70% of the time.
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