Question

There are four washing machines in an apartment complex: A, B, C, D. On any given day the probability that these machines break down is as follows:
P(A) = 0.04, P(B) = 0.01, P(C) = 0.06, P(D) = 0.01 .
Assume that the functionality of each machine is independent of that of others. What is the probability that on a given day at least one machine will be working?

Answer 1

Let A be first event that the washing machine not working, B be the second event that the washing machine not working, C be third event that the washing machine not working and D be fourth event that the washing machine not working.
From the given information, the P (4) = 0.04, P (B) =0.01, P (C) =0.06, P (D)=0.01 and all events are independent.
Step 2
Multiplication rule for independent events:
\(\displaystyle{P}{\left({A}\cap{B}\cap{C}\cap{D}\right)}={P}{\left({A}\right)}\times{P}{\left({B}\right)}\times{P}{\left({C}\right)}\times{P}{\left({D}\right)}\)
Then, the probability that at least one machine will be working is
P(At least one machine working)=1-P(None of them working)
\(\displaystyle={1}-{P}{\left({A}\cap{B}\cap{C}\cap{D}\right)}\)
\(\displaystyle={1}-{P}{\left({A}\right)}\times{P}{\left({B}\right)}\times{P}{\left({C}\right)}\times{P}{\left({D}\right)}\)
\(\displaystyle{\left(\therefore\ \text{ Events are independent}\right)}\)
\(\displaystyle={1}-{\left({0.04}\times{0.01}\times{0.06}\times{0.01}\right)}\)
=1-0.00000024
=0.99999976
Thus, the probability that at least one machine will be working is 0.99999976 which is approximately 1