Answer:
(a) The probability that an item selected for inspection is classified as defective is 0.01189.
(b) The probability that an item selected at random is classified as non-defective when in fact it is good is 0.99992.
Step-by-step explanation:
Let's denote the events as follows:
G = an item is good
B = an item is bad
D = an item is classified as defective.
Given:
[tex]P (D|B) = 0.99\\P(D|G)=0.005\\P(B)=0.007[/tex]
The probability of producing good items is:
[tex]P(G)=1-P(B)=1-0.007=0.993[/tex]
(a)
The law of total probability states that:
[tex]P(A)=P(A|B)P(B)+P(A|C)P(C)[/tex]
Using the law of total probability determine the probability that an item selected for inspection is classified as defective as follows:
[tex]P(D)=P(D|G)P(G)+P(D|B)P(B)\\=(0.005\times0.993)+(0.99\times0.007)\\=0.01189[/tex]
Thus, the probability that an item selected for inspection is classified as defective is 0.01189.
(b)
Compute the probability that an item selected at random is classified as non-defective when in fact it is good as follows:
[tex]P(G|D^{c})=\frac{P(D^{c}|G)P(G)}{P(D^{c})} \\=\frac{[1-P(D|G)]P(G)}{1-P(D)} \\=\frac{[1-0.005]\times0.993}{1-0.01189} \\=0.99992[/tex]
Thus, the probability that an item selected at random is classified as non-defective when in fact it is good is 0.99992.
