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

(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.

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)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)=(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* 0.08)+(0.16* (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)=(P(Y|X)P(X))/(P(Y))`

`=(0.64* 0.08)/(0.1984)\\\\=0.2581`

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