Question

Two method, A and B, are available for teaching a certain industrial skill. The failure rate is 20% for method A and 10% for method B. Method B is more expensive, however, and hence is used only 40% of the time. (Method A is used the other 60% of the time.) A worker is taught the skill by one of the two methods, but he fails to learn it correctly. What is the probability that he was taught by using method A? Show all work.

(A) What is the probability that the worker fails to learn it correctly? Include set notation or use a tree diagram.

(B) What is the probability that he was taught by using method A if he failed to learn it correctly? Use Bayes’ rule.

Answer 1

A : Event that worker was taught by using method A

P(A) = 60/100 =0.6

B : Event that worker was taught by using method B

P(B) = 40/100 =0.4

(A)

F : Event that the worker fails to learn it correctly.

The failure rate is 20% for method A

i.e

Probability that the the worker fails to learn it correctly given that that worker was taught by using method A = P(F|A) =20/100 =0.20

The failure rate is 10% for method B

i.e

Probability that the the worker fails to learn it correctly given that that worker was taught by using method B = P(F|B) =10/100 =0.10

F : Event that the worker fails to learn it correctly.

Event that the worker fails to learn it correctly is equivalent :

Event that worker was taught by using method A and he fails to learn it correctly OR

Event that worker was taught by using method A and he fails to learn it correctly

i.e

F = (A and F) OR (B and F)

Probability that the worker fails to learn it correctly : P(F)

P(F) = P(A and F) + P(B and F) = P(A)P(F|A) + P(B) P(F|B)

P(A)P(F|A) = 0.60 x 0.20 = 0.12

P(B)P(F|B) = 0.4 x 0.10 = 0.04

P(F) = P(A)P(F|A) + P(B) P(F|B) = 0.12+0.04 = 0.16

Probability that the worker fails to learn it correctly = 0.16

B) Probability that he was taught by using method A if he failed to learn it correctly

Probability that he was taught by using method A if he failed to learn it correctly = P(A|F)

By Bayes theorem,

P(A)P(F|A) = 0.60 x 0.20 = 0.12

P(B)P(F|B) = 0.4 x 0.10 = 0.04

P(A)P(F|A) + P(B) P(F|B) = 0.12+0.04 = 0.16

Probability that he was taught by using method A if he failed to learn it correctly = 0.75