Question

An electric component manufactured by 'RASU Electronics' is tested for its defectiveness by a sophisticated testing device. Let A denote the event "the device is defective " and B the event "the
testing device reveals the component to be defective ". Suppose \[P\left( A \right)=\alpha \] and
\[P(B\left| A \right.)=P(B'\left| A \right.')=1-\alpha \] where \[0<\alpha <1\], then
A. \[P\left( B \right)=2\alpha \left( 1-\alpha \right)\]
B. \[P\left( A\left| B' \right. \right)=\dfrac{1}{2}\]
C. \[P\left( B' \right)={{\left( 1-\alpha \right)}^{2}}\]
D. \[P\left( A'\left| B' \right. \right)={{\left[ \alpha /\left( 1-\alpha \right) \right]}^{2}}\]

Answer 1

Hint:First we will find the value of the probability of B from the given probability (conditional) as the formula for the mathematical behaviors of such probability is given as:

\[P\left( A\left| B \right. \right)=\dfrac{P\left( A,B \right)}{P\left( A \right)}\]
And the probability to find device being defective but not able to find it as:

\[P\left( A\left| B' \right. \right)=\dfrac{P\left( A\cap B' \right)}{1-P\left( B \right)}\]

Hence, using the above two probability methods we will find which of the options are correct and which are not.

Complete step by step solution:
With the information \[P\left( A \right)=\alpha \] and \[P(B\left| A \right.)=P(B'\left| A \right.')=1-\alpha
\] given, the probability \[P(A)\] is denoted as the event that proves whether the device is defective or not, and \[P(B)\] is denoted as the testing device being defective.

Now using the data given in the question, the probability of the device being defective is written as:
\[\Rightarrow P(B)=P\left( A \right)P\left( B\left| A \right. \right)+P\left( A' \right)P\left( B\left| A' \right.
\right)\]

Replacing the value of \[P\left( B\left| A' \right. \right)\] by \[\left( 1-P\left( B'\left| A' \right. \right)
\right)\] as both the probability of device being defective and not finding one.
\[\Rightarrow P(B)=P\left( A \right)P\left( B\left| A \right. \right)+P\left( A' \right)\left( 1-P\left( B'\left|
A' \right. \right) \right)\]

Replacing the values by \[P\left( A \right)=\alpha \] and \[P(B\left| A \right.)=P(B'\left| A \right.')=1-
\alpha \], we get the value of:
\[\Rightarrow \alpha \left( 1-\alpha \right)+\left( 1-\alpha \right)\left[ 1-\left( 1-\alpha \right) \right]\]
\[\Rightarrow 2\alpha \left( 1-\alpha \right)\]

Now using the formula for the probability when the device is defective but still passes the inspection:

\[\Rightarrow P\left( A\left| B' \right. \right)=\dfrac{P\left( A\cap B \right)}{P\left( B \right)}\]
\[\Rightarrow P\left( A\left| B' \right. \right)=\dfrac{P\left( B \right)-P\left( A\cap B \right)}{P\left( B
\right)}\]
\[\Rightarrow P\left( A\left| B' \right. \right)=\dfrac{P\left( B \right)-P\left( A \right)P\left( B\left| A
\right. \right)}{P\left( B \right)}\]

Replacing the values by \[P\left( A \right)=\alpha \] and \[P(B\left| A \right.)=P(B'\left| A \right.')=1-
\alpha \], we get the value of:
\[\Rightarrow P\left( A\left| B' \right. \right)=\dfrac{2\alpha \left( 1-\alpha \right)-\alpha \left( 1-\alpha
\right)}{2\alpha \left( 1-\alpha \right)}\]
\[\Rightarrow P\left( A\left| B' \right. \right)=\dfrac{1}{2}\]

Now finding the probability when both the device is not defective and is not marked defective during inspection as well is:

\[P\left( A'\left| B' \right. \right)=\dfrac{P\left( A'\left| B' \right. \right)}{P\left( B' \right)}\]

Replacing the value in terms of its opposite condition, we get the probability as:
\[\Rightarrow P\left( A'\left| B' \right. \right)=\dfrac{1-P\left( A\cup B \right)}{1-P\left( B \right)}\]
\[\Rightarrow P\left( A'\left| B' \right. \right)=\dfrac{P\left( A \right)+P\left( B \right)-P\left( A\cap B
\right)}{1-P\left( B \right)}\]
\[\Rightarrow P\left( A'\left| B' \right. \right)={{\left( \dfrac{\alpha }{1-\alpha } \right)}^{2}}\]

Hence, the options \[P\left( B \right)=2\alpha \left( 1-\alpha \right)\], \[P\left( A\left| B' \right.
\right)=\dfrac{1}{2}\] and \[P\left( A'\left| B' \right. \right)={{\left[ \alpha /\left( 1-\alpha \right)
\right]}^{2}}\] are all correct.

Note:The formula is coming from the derivation of the Bayes’s theorem where A and B are two events and \[P\left( A\left| B \right. \right)\] proves that probability of A is given and the probability of B is true and vice versa for \[P\left( B\left| A \right. \right)\] and probability of event A happening is \[P(A)\] and B is \[P(B)\] and all of them are combined as:
\[\dfrac{P\left( A\left| B \right. \right)}{P\left( B\left| A \right. \right)}=\dfrac{P\left( A \right)}{P\left( B \right)}\]