Faulty Inspection

• assuming 1000 items in total.
• no. of defective = 10% * 1000 = 100
• no. of inspected from defective pile = 60% * 100 = 60
• no. of inspected from good pile = 20% * 900 = 180
• no. of inspected = 60 + 180 = 240
• likelihood of inspected is defective = 60/240 = 0.25

Let the events $D$ and $I$ be where a product is defective and inspected, respectively. For defective items, we are given that $P(D) = 0.1$ and $P(I|D) = 0.6$ . For good items, we are given $P(\neg D) = 1 - P(D) = 0.9$ and $P(I|\neg D) = 0.2$ .

From the definition of conditional probability and since each product is either defective or not defective, we find $P(I)$ :

$P(I) = P(D)P(I|D) + P(\neg D)P(I|\neg D) = 0.1 \cdot 0.6 + 0.9 \cdot 0.2 = 0.24$ .

Now, applying Bayes' Rule:

$P(D|I) = \frac{P(I|D)P(D)}{P(I)} = \frac{0.1(0.6)}{0.24} = 0.25$ .

So, $P(D|I) = 0.25$ .