SOLUTION: 2. A machine has twelve identical components that function independently. It will stop working if three or more of the components fail. If the probability that an individual com
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Question 346244: 2. A machine has twelve 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 0.11, find the probability that the machine will be working.
Found 2 solutions by Fombitz, solver91311:
Answer by Fombitz(32388) (Show Source): You can put this solution on YOUR website!
Look at the probability of , , and components failing.
failures:
failure:
failures:
.
.
.
P(working)=
.
.
.
The coefficients are generated by which are the first 3 Pascal triangle coefficients for () terms.
Answer by solver91311(24713) (Show Source): You can put this solution on YOUR website!
The probability of 
successes in 
trials where 
is the probability of success on any given trial is given by:
\ =\ \left(n\cr k\right\)p^k\left(1\,-\,p\right)^{n\,-\,k})
Where )
is the number of combinations of things taken at a time and is calculated by !})
But what you need is the probability that there are no more than 2 failed components. Hence you need the probability of exactly 0 plus the probability of exactly 1 plus the probability of exactly 2.
\ =\ \sum_{i=0}^2\left(11\cr\ i\right\)\left(0.11\right)^i\left(0.89\right)^{n\,-\,i})
Which is to say:
\ =\ P_{11}(0,0.11)\ +\ P_{11}(1,0.11)\ + P_{11}(2,0.11))
\left(0.11\right)^0\left(0.89\right)^{11}\ +\ \left(11\cr\, 1\right\)\left(0.11\right)^1\left(0.89\right)^{10}\ +\ \left(11\cr\, 2\right\)\left(0.11\right)^2\left(0.89\right)^{9})
\left(1\right)\left(0.89\right)^{11}\ +\ \left(11\right\)\left(0.11\right)^1\left(0.89\right)^{10}\ +\ \left(55\right\)\left(0.11\right)^2\left(0.89\right)^{9})
I'll leave you alone to spend some nice quality time with your calculator.
John
My calculator said it, I believe it, that settles it
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