Question

A system has 30 independent circuits each ciruit has 0.015 odds of being faulty. The system is operational while there are at most 2 faulty circuits. What is the probability that he system is operational?

Answer 1

Since the events are independent of each other and  fixed number of trials with two outcomes it is a binomial probability distribution

Binomial probability is given by

P(X=x) = C(n,x)*px*(1-p)(n-x)

Sample size , n =    30
Probability of an fault, p =   0.015

Now system will be operational when there are at most 2 faulty circuits which means  

Probability of operation = P(0) + P(1) +P(2)

P ( X =    0   ) = C(30,0) * 0.015^0 * (1-0.015)^30 =            0.63546

P ( X =    1   ) = C(30,1) * 0.015^1 * (1-0.015)^29 =            0.29031

P ( X =    2   ) = C(30,2) * 0.015^2 * (1-0.015)^28 =            0.06410

Probability of operation =  0.63546 + 0.29031 + 0.06410 =  0.98987

Please revert back in case of any doubt.