Question

The following circuit operates if and only there is a path of functional devices from left to right.
the probability that each device functions is as shown.
Assume that the probability that a device functions does not depend on whether or not other devices are functional. what is the probability that the circuit operates.

Answer 1

Concepts and reason

Properties of probability is used to obtain the results of this questions. Different properties of probability are applied in different parts of this question. Property of compliment, different cases are made as per different requirements of the questions.

Fundamentals

The probability is likelihood of happening of a particular event. The event can be anything justified. The sum of the probability of each of the event in a particular sample space would be equal to one. This concept can be used to obtain the probability of non-happening an event provided that probability of happening of an event is provided. If two events are independent to each other, then the probability of happening of both events can be obtaining by multiplying their respective probability.

The sum of probability of two complementary events would always be equal to one. Hence,

$P\left( {\overline A } \right) + P\left( A \right) = 1$

If two events $E$ and $F$ are independent random variable, then the probability of $E \cap F$ can be obtained as follows:

$P\left( {E \cap F} \right) = P\left( E \right)P\left( F \right)$

The probability that the circuit would work if any of the two path works. It can be obtained as follows:

$P\left( {{\rm{either}}\,{\rm{of}}\,{\rm{the}}\,{\rm{path}}\,{\rm{works}}} \right) = 1 - p\left( {{\rm{both}}\,{\rm{do}}\,{\rm{not}}\,{\rm{work}}} \right)$

The probability that first path works can be obtained as follows:

‎ $\begin{array}{c}\\P\left( {{\rm{First}}\,{\rm{Path}}\,{\rm{Works}}} \right) = 0.9 \times 0.8 \times 0.7\\\\ = 0.504\\\end{array}$

The probability that the second path works can be obtained as follows:

$\begin{array}{c}\\P\left( {{\rm{Second}}\,{\rm{Path}}\,{\rm{Works}}} \right) = {0.95^3}\\\\ = 0.8573\\\end{array}$

Using the probabilities obtained, the probability that both of the path does not work can be obtained as follows:

$\begin{array}{c}\\P\left( {{\rm{both}}\,{\rm{do}}\,{\rm{not}}\,{\rm{work}}} \right) = \left( {1 - 0.504} \right)\left( {1 - 0.8573} \right)\\\\ = 0.070742\\\end{array}$

The required probability can be obtained as follows:

$\begin{array}{c}\\P\left( {{\rm{either}}\,{\rm{of}}\,{\rm{the}}\,{\rm{path}}\,{\rm{works}}} \right) = 1 - p\left( {{\rm{both}}\,{\rm{do}}\,{\rm{not}}\,{\rm{work}}} \right)\\\\ = 1 - 0.070742\\\\ = 0.929258\\\end{array}$

Ans:

The probability that the circuit functions would be $0.929258$ .