The machine will be working if
1. no part fails ---> (.8)^7
2. 1 part fails ----> C(7,1)(.2)(.8)^6
3. 2 parts fail ---> C(7,2)(.2^2)(.8^5)
4. 3 parts fail ---> C(7,3)(.2^3)(.8^4)
add them up
A machine has 7 identical components which function independently. The probability that a component will fail is .2. The machine will stop working if more than three components fail. Find the probability that the machine will be working?
2 answers
.966