Answer to Question #245569 in Statistics and Probability for Fire

Question

The system works if components A and B work or components C, D, and E work. Assume that the components fail independently with the following probabilities: P(A fails) = P(B fails) = 0.1 and P(C fails) = P(D fails) = P(E fails) = 0.2.

(a) What is the probability that the system works?

(b) Given that the system works, what is the probability that component A does not work? (c) Given the system does not work, what is the probability that component A also does not work?

Answer 1

Since the events are independent, probabilities that the components work are given as,

P(components AB works)=P(A AND B works)=(1-0.1)²=0.9²=0.81

P(components CDE works)=P(C AND D AND E works)=(1-0.2)³=0.8³=0.512

a.

Probability that the system works is given as,

P(system works)=1-P(system does not work)

=1-((1-0.81)*(1-0.512))

=1-(0.19*0.488)

=1-0.09272

=0.9073(4 decimal places)

Therefore, probability that the system works is 0.9073.

b.

Probability that A does not work given that the system works is given as,

P(A does not work| system works) =P(A does not work AND CDE works)/P(system works)

=(0.1*0.512)/0.9073

=0.0564(4 decimal places).

Hence, probability that A does not work given the system works is 0.0564.

c.

Probability that A does not work given that the system does not work is given by,

P(A does not work| system does not work)=

P(A does not work AND CDE does not work)/P(system does not work)=

(0.1)*(1-0.512)/(1-0.9073)=(0.1*0.488)/0.0927=0.5263( 4 decimal places)

Thus, Probability that A does not work given that the system does not work is 0.5263.