Question

In a manufacturing process where 8% of all items are defective, an inspector at the end of the production line selects some items for inspection. Suppose that 64% of the defective items and 16% of the good items go through the inspection. Find the probability that an item

a. has an inspection
b. is defective given that it has an inspection.

Answer 1

Answer:

(a) The probability that an item  has an inspection is 0.1984.

(b) The probability that an item is defective given that it has an inspection is 0.2581.

Step-by-step explanation:

The law of total probability states that, if B₁, B₂, B₃,... are part of a sample space S, then for any event A,

$P(A)=\sum\limits_{i}{P(A|B_{i})\cdot P(B_{i})}$

The condition probability of an event A given that another event B has already occurred is:

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

Denote the events as follows:

X = an item is defective

Y = the item is being inspected by the inspector.

The information provided is:

$P (X) = 0.08\\P (Y |X)=0.64\\P(Y|X^{c})=0.16$

(a)

Compute the probability that an item  has an inspection as follows:

Use the law of total probability:

$P(Y)=P(Y|X)P(X)+P(Y|X^{c})P(X^{c})$

         $=(0.64\times 0.08)+(0.16\times (1-0.08))\\=0.0512+0.1472\\=0.1984$

Thus, the probability that an item  has an inspection is 0.1984.

(b)

Compute the probability that an item is defective given that it has an inspection as follows:

Use the condition probability:

$P(X|Y)=\frac{P(Y|X)P(X)}{P(Y)}$

             $=\frac{0.64\times 0.08}{0.1984}\\\\=0.2581$

Thus, the probability that an item is defective given that it has an inspection is 0.2581.