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Question: An electric component manufactured by 'RASU Electronics' is tested for its defectiveness by a sophis...

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(A)=αP\left( A \right)=\alpha and
P(B∣A)=P(B′∣A′)=1−αP(B\left| A \right.)=P(B'\left| A \right.')=1-\alpha where 0<α<10<\alpha <1, then
A. P(B)=2α(1−α)P\left( B \right)=2\alpha \left( 1-\alpha \right)
B. P(A∣B′)=12P\left( A\left| B' \right. \right)=\dfrac{1}{2}
C. P(B′)=(1−α)2P\left( B' \right)={{\left( 1-\alpha \right)}^{2}}
D. P(A′∣B′)=[α/(1−α)]2P\left( A'\left| B' \right. \right)={{\left[ \alpha /\left( 1-\alpha \right) \right]}^{2}}

Explanation

Solution

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(A∣B)=P(A,B)P(A)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(A∣B′)=P(A∩B′)1−P(B)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(A)=αP\left( A \right)=\alpha and P(B∣A)=P(B′∣A′)=1−αP(B\left| A \right.)=P(B'\left| A \right.')=1-\alpha given, the probability P(A)P(A) is denoted as the event that proves whether the device is defective or not, and P(B)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:

\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)}$$