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

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