MA305 Mathematics for ICE 1

MA305Conditional Probability

Bayes’ TheoremBy: Prof. Nutan Patel

Asst. Professor in MathematicsIT-NUA-203

patelnutan.wordpress.com

MA305 Mathematics for ICE 2

Conditional Probability• Definition: If A and B are two events associated with the same 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:

0)(,)|(

BPBP

BAPBAP

)|(1)|(.1 AAPAUP .),|()()|()()(.2 ruletionmultiplicaGeneralBAPBPABPAPBAP

).()|()()|(,.3 BPABPandAPBAPEventstIndependenFor

MA305 Mathematics for ICE 3

General Multiplication Rule If A and B are any events in S, then P(AB) =P(A)P(B|A), if P(A) ≠ 0

=P(B)P(A|B), if P(B) ≠ 0.

Ex- Two cards are drawn at random from an ordinary deck of 52 playing cards. What is the probability of getting two aces if(a) The first card is replaced before the second card is drawn;(b) The first card is not replaced before the second card is drawn?

Ans: Let event A be first card is ace and B be second card is ace.(c) Since there are four aces among the 52 playing cards, we get P(AB)= =.(b) P(AB) =P(A)P(B|A)= =.

MA305 Mathematics for ICE 4

Ex-2. The supervisor of a group of 20 workers wants to get the opinion of 2 of them (to be selected random) about certain new safety regulations. If 12 of them favour the new regulations and the other 8 are against it, what is the probability that both of the workers chosen by the supervisor will be against safety regulations?

Ans- the probability that the first worker selected will be against the new safety regulations is P(A)=, and the probability that the second worker selected will be against the new safety regulations given that the first one is against them is P(B|A)=Thus, the desired probability is P(AB)=P(A)P(B|A)==.

MA305 Mathematics for ICE 5

Ex-3. Among 60 automobile repair parts loaded on a truck in Bombay, 45 are destined for Vadodara and 15 for Ahmedabad. If two of parts are unloaded in Surat by mistake and the “selection” is random, what are the probabilities that(a) Both parts should have gone to Vadodara;(b) Both parts should have gone to Ahmedabad;(c) One should have gone to Vadodara and one to Ahmedabad.

Ans- (a) = =0.5593 (b) = = 0.059322

(c) 1-(+)=0.381 or

== 0.381

MA305 Mathematics for ICE 6

0.24 0.190.06

0.110.16

0.04

Ex-4. With reference to Figure, find(a) P(A|B)(b) P(B|C’)(c) P(AB|C)(d) P(BC|A’)(e) P(A|BC)(f) P(A|BC)(g) P(ABC|BC)(h) P(ABC|BC)

Ans: (a) 0.25 (b) 0.417 (c) 0.1 (d) 0.78 (e) 0.4 (f) 0.267(g) 0.267 (h) 0.062

A B

C0.11

U

MA305 Mathematics for ICE 7

•Rule of total probabilityExample: Suppose that an assembly plant receives its voltage regulators from three different suppliers, 60% from supplier B1, 30% from supplier B2, and 10 % from supplier B3. If 95% of the voltage regulators from B1, 80% of those from B2, and 65% of those from B3 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 B1, B2 and B3 are the events that it comes from respective suppliers,

We can write A = A(B1 B2 B3 )

= (A B1)(A B2)(A B3)

and P(A) = P((A B1)(A B2)(A B3))

= P(A B1)+P(A B2)+P(A B3) (as B1 B2 B3 =)

= P(B1)P(A| B1)+P(B2)P(A| B2)+P(B3)P(A| B3)= (0.60)(0.95)+(0.30)(0.80)+(0.10)(0.65)= 0.875

MA305 Mathematics for ICE 8

Rule of total probabilityIf B1, B2, … , Bn are mutually exclusive events of which one must occur, then

P(A)=Ex: (continue)

Suppose we want to know the probability that a particular voltage regulator perform according to specifications, came from supplier B3.

i.e. P(B3|A)We write

=

==0.074.

Note that the probability that a voltage regulator is supplied by B3 decreases from 0.10 to 0.074 once it is known that it perform according to specifications.

MA305 Mathematics for ICE 9

Bayes’ TheoremIf B1, B2, … , Bn are mutually exclusive events of which one must occur, then

, for m=1, 2, … , n.

Bayes’ theorem provides a formula for finding the probability that the “effect” A was “caused” by event Bm.

MA305 Mathematics for ICE 10

• By Tree Diagram

B1

B2

B3

A

A

A

P(A| B1)

P(A| B2)

P(A| B3)

P(B 1)

P(B3 )

P(B2)

Total Probability: P(A)= P(B1)P(A| B1)+P(B2)P(A| B2)+P(B3)P(A| B3)

Bayes’ Theorem: =

MA305 Mathematics for ICE 11

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 defectivei. 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)=0.037i. P(X|A)= 0.2703 ii. P(Y|A)=0.4054 iii. P(Z|A)=0.3243

MA305 Mathematics for ICE 12

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: 0.0684iv. 0.2923v. 0.3508vi. 0.3508

MA305 Mathematics for ICE 13

• 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