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 $P\left(B|A\right)=P\left(B^{\prime }|A^{\prime }\right)=1-\alpha$, where $0<\alpha <1$, then

• A. $P\left(B\right)=2\alpha (1-\alpha )$
• B. $P\left(A|B^{\prime }\right)=1/2$
• C. $P\left(B^{\prime }\right)=(1-\alpha {)}^2$
• D. $P\left(A^{\prime }|B^{\prime }\right)=\left[\alpha /\right(1-\alpha ){]}^2$

Answer 1

Answer: (A), (B), (C)

The correct options are
A $P\left(B\right)=2\alpha (1-\alpha )$
B $P\left(A|B^{\prime }\right)=1/2$
C $P\left(B^{\prime }\right)=(1-\alpha {)}^2$

$P\left(B\right)=P\left(A\right)P\left(B|A\right)+P\left(A^{\prime }\right)P\left(B|A^{\prime }\right)$
$=P\left(A\right)P\left(B|A\right)+P\left(A^{\prime }\right)[1-P(B^{\prime }|A^{\prime }\left)\right]$
$=\alpha (1-\alpha )+(1-\alpha )[1-(1-\alpha \left)\right]=2\alpha (1-\alpha )$

and $P\left(A|B^{\prime }\right)=\frac{P(A/\cap B)}{P\left(B\right)}=\frac{P\left(B\right)-P(A\cap B)}{P\left(B\right)}$

$=\frac{P\left(B\right)-P\left(A\right)P\left(B|A\right)}{P\left(B\right)}$

$=\frac{2\alpha (1-\alpha )-\alpha (1-\alpha )}{2\alpha (1-\alpha )}=\frac{1}{2}$

$P\left(A^{\prime }|B^{\prime }\right)=\frac{P\left(A^{\prime }|B^{\prime }\right)}{P\left(B^{\prime }\right)}=\frac{1-P(A\cup B)}{1-P\left(B\right)}$

and $P(A\cup B)=P\left(A\right)+P\left(B\right)-P(A\cap B)={\alpha }^2$