~Advanced probability techniques~

~Consider a universe with 3 mutually exclusive (cannot happen at the same time) events, A(1), A(2), & A(3).

~Let the union of these events make up the entire sample space, S.

~Let event E be a subset of the sample space S that shares an intersection (points in common) with all 3 events.

~Then, P(E)=P(E & A1) + P(E & A2) + P(E & A3)  

~Now multiply the top and bottom of each term on the right side by the indicated quantity. (this does not change the value of the expression).

~P(E)=P(A1)P(E & A1)/P(A1) + P(A2)P(E & A2)/P(A2) + P(A3)P(E & A3)/P(A3)

~Then, (1) P(E)=P(A1)P(E|A1) + P(A2)P(E|A2)+P(A3)P(E|A3),
since, P(A|B)=P(A and B)/P(B)

~Note:  Equation (1) is known as the Law of total probability & can be extended to k cases.

~Example:  Companies A, B and C produce laptop computers. Company A has 20% of them on the shelves of store X, B has 30%, and C has 50%. The defect rates for production are 2% for A, 3% for B and 5% for C. Find the probability that when a consumer buys one at store X it has a defect.

~Note in equation (1) above, E=D=defect, A1=produced at A, A2=produced at B, A3=produced at C.

~So, P(D)=P(produced at A)P(defect from A) + P(produced at B)P(defect from B) + P(produced at C)P(defect from C) = (.20)(.02) + (.30)(.03) + (.50)(.05) = .038 = 3.8%

~Therefore, the chance that the consumer buys a defective laptop at X is 3.8%

~Let’s look at P(A2|E) or any other one where A is first.

~This translates our example problem to finding the probability that the laptop came from company B, given the fact that it has a defect. ( this is different from what we did above).

~(2) P(A2|E)=P(A2 & E)/P(E)=P(A2 & E) divided by [P(A1)P(E|A1)+P(A2)P(E|A2)+P(A3)P(E|A3)], (just replace P(E) by equation (1), in the denominator).

~Note: This is a form of Bayes’formula.

~Remember, these formulas can be extended to k cases (i.e., A1,A2,A3,A4, ... Ak).

~Example:  Using the previous example, What is the probability that the defected laptop purchased from store X came from company B. Here, E=D and A2=B, So, we have, P(came from B given that it’s defective) = P(B|D) = P(B and D)/P(D) = (.30)(.03)/.038 ←( .038, from the result of above). Which gives .237 = 23.7% chance that the defective laptop purchased came from B.

~These are interesting formulas that most students exposed to probability do not know & are useful in solving many problems of conditional probability.