Question:
Denote A, B as two random events. Given P(A) = 1/4, P(B|A) = 1/3, P(A|B) = 1/2, compute P(\overline{A}\text{ }\overline{B}).
Solution:
By Multiplication Rules:
P(AB) = P(A)P(B|A) = (1/4)‧(1/3) = 1/12
P(B) = \frac{P(AB)}{P(A|B)} = \frac{1/12}{1/2} == 1/6
By Subtraction Rules:
P(\overline{A}B)= P(B) - P(AB) = 1/6 - 1/12 = 1/12
P(\overline{A}\text{ }\overline{B}) = P(\overline{A}) - P(\overline{A}B)
= [1-P(A)] - P(\overline{A}B) = (1 - 1/4) - 1/12 = 2/3.
Question:
There are 10 products in each box, in which the probabilities of taking 0, 1 and 2 bogus products are the same. When performing open-box quality check (QC) by picking up a product randomly, if the product is bogus, the QC will fail and the box will be rejected. Since there are errors in QC, the probability of a genuine product being mistaken as bogus is 2%, while the probability of a bogus product being mistaken as genuine is 10%. For a random QC on a box, what is the probability p of passing the QC?
Solution:
Denote event A as "The product passes the QC", then P(A) = p. Also, denote event B as "picking up any one genuine product, and \overline{B} as "picking up any one bogus product. Then we have:
A = BA \cup \overline{B} A
Given the following conditional probabilities:
P(A|B) = 1 - 0.02 = 0.98
P(A|\overline{B}) = 0.1
So:
p = P(A) = P(A|B) + P(A|\overline{B})
= P(B)P(A|B) + P(\overline{B})P(A|\overline{B})
= 0.98P(B) + 0.1[1-P(B)] = 0.1 + 0.88P(B)
Note that P(B) is related to several instance of picking up bogus products in a box, so we will apply Total Probability Theorem to compute P(B):
Denote C_i as "There are i amount(s) of bogus products in each box", then C_0, C_1, C_2 are collectively exhaustive events. Now:
P(C_i) = 1/3
B = C_0B \cup C_1B \cup C_2B
So:
P(B) = P(C_0)P(B|C_0) + P(C_1)P(B|C_1) + P(C_2)P(B|C_2)
= (1/3)(10/10) + (1/3)(9/10) + (1/3)(8/10) = 0.9
Then:
p = 0.1 + (0.88)(0.9) = 0.892.