Mathematics: 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.

It makes sense. hope this even though someone already answered it. : p...

In a manufacturing process where 8% of all items

staciagirl3880

26.01.2022

(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.

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