MA 305 Conditional Probability Bayes Theorem By Prof

MA 305 Conditional Probability Bayes’ Theorem By: Prof. Nutan Patel Asst. Professor in Mathematics IT-NU A-203 patelnutan. wordpress. com MA 305 Mathematics for ICE 1

Conditional Probability • Definition: If A and B are two events associated with the sample space of a random experiment, the conditional probability of the event A given that B has occurred, i. e. P (A|B) is given by Properties: MA 305 Mathematics for ICE 2

• MA 305 Mathematics for ICE 3

• MA 305 Mathematics for ICE 4

• MA 305 Mathematics for ICE 5

Ex-4. With reference to Figure, find (a) P(A|B) (b) P(B|C’) (c) P(A B|C) (d) P(B C|A’) (e) P(A|B C) (f) P(A|B C) (g) P(A B C|B C) (h) P(A B C|B C) Ans: (a) 0. 25 (d) 0. 78 (g) 0. 267 (b) 0. 417 (e) 0. 4 (h) 0. 062 U A B 0. 24 0. 06 0. 19 0. 04 0. 16 0. 11 0. 09 C (c) 0. 1 (f) 0. 267 MA 305 Mathematics for ICE 6

• Rule of total probability Example: Suppose that an assembly plant receives its voltage regulators from three different suppliers, 60% from supplier B 1, 30% from supplier B 2, and 10 % from supplier B 3. If 95% of the voltage regulators from B 1, 80% of those from B 2, and 65% of those from B 3 perform according to specifications, what is the probability that any one voltage regulator received by the plant will perform according to specifications. Ans: If A denotes the event that a voltage regulator received by the plant performs according to specifications, and B 1, B 2 and B 3 are the events that it comes from respective suppliers, We can write A = A (B 1 B 2 B 3 ) = (A B 1) (A B 2) (A B 3) and P(A) = P((A B 1) (A B 2) (A B 3)) = P(A B 1)+P(A B 2)+P(A B 3) (as B 1 B 2 B 3 = ) = P(B 1)P(A| B 1)+P(B 2)P(A| B 2)+P(B 3)P(A| B 3) = (0. 60)(0. 95)+(0. 30)(0. 80)+(0. 10)(0. 65) = 0. 875 MA 305 Mathematics for ICE 7

• MA 305 Mathematics for ICE 8

Bayes’ Theorem • MA 305 Mathematics for ICE 9

P( • By Tree Diagram B 1) P(B 2) P(B 3) B 1 B 2 B 3 P(A| B 1) P(A| B 2) P(A| B 3) A A A MA 305 Mathematics for ICE 10

Example: A Manufacturer buys an item from three subcontractors, X, Y ans Z. X has the better quality control, only 2% of its item are defective. X furnishes the manufacturer with 50% of items. Y furnishes 30% of the items, and 5% of its items are defective. Z furnishes 20% of the items, and 6% of its items are defective. The manufacturer finds an item defective i. What is the probability that it came from X? ii. What is the probability that it came from Y? iii. What is the probability that it came from Z? Ans: P(A)=0. 037 i. P(X|A)= 0. 2703 ii. P(Y|A)=0. 4054 iii. P(Z|A)=0. 3243 MA 305 Mathematics for ICE 11

Example: A consulting firm rents cars from three agencies, 20% from agency D, 20% from agency E, and 60% from agency F. if 10% of the cars from D, 12% of the cars from E, and 4% of the cars from F have bad tires, what is the probability that the firm will get a car with bad tires? i. What is the probability that bad tires came from D? ii. What is the probability that bad tires came from E? iii. What is the probability that bad tires from F? Ans: 0. 0684 i. 0. 2923 ii. 0. 3508 iii. 0. 3508 MA 305 Mathematics for ICE 12

• A Company manufactures integrated circuits on silicon chips at three different plants X, Y, and Z. out of every 1000 chips produced, 400 come from X, 350 come from Y, and 250 come from Z. it has been estimates that of the 400 from X 10 are defective, wheres five of those from Y are defective, and only two of those from Z are defective. Determine the probability that a defective chip came from plant Y. • Ans: 0. 324 MA 305 Mathematics for ICE 13