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(C) Let $D$ be the event that the item is defective and $ND$ be the event that the item is not defective. Given $P(D) = 0.3$,so $P(ND) = 1 - 0.3 = 0.7

Vedclass · Accepted Answer

$\frac{3}{31}$

Vedclass · Answer

(C) Let $D$ be the event that the item is defective and $ND$ be the event that the item is not defective. Given $P(D) = 0.3$,so $P(ND) = 1 - 0.3 = 0.7$. Let $R_D$ be the event that the device reports the item as defective and $R_{ND}$ be the event that the device reports the item as not defective. The device is accurate $80\%$ of the time,so $P(R_D|D) = 0.8$ and $P(R_{ND}|ND) = 0.8$. Consequently,$P(R_{ND}|D) = 1 - 0.8 = 0.2$ and $P(R_D|ND) = 1 - 0.8 = 0.2$. We need to find the probability that the item is defective given that the device reports it as not defective,i.e.,$P(D|R_{ND})$. Using Bayes' Theorem: $P(D|R_{ND}) = \frac{P(R_{ND}|D) 	imes P(D)}{P(R_{ND}|D) 	imes P(D) + P(R_{ND}|ND) 	imes P(ND)}$ $P(D|R_{ND}) = \frac{0.2 	imes 0.3}{(0.2 	imes 0.3) + (0.8 	imes 0.7)}$ $P(D|R_{ND}) = \frac{0.06}{0.06 + 0.56} = \frac{0.06}{0.62} = \frac{6}{62} = \frac{3}{31}$.

An item is tested on a device for its defectiveness. The probability that such an item is defective is $0.3$. The device gives an accurate result in $8$ out of $10$ such tests. If the device reports that an item tested is not defective,then the probability that it is actually defective is:

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