Question

Two methods, A and B, are available for teaching a certain industrial skill. The failure rate is 15% for A and 5% for B. However, B is more expensive and hence is used only 40% of the time. (A is used the other 60%.) A worker is taught the skill by one of the methods but fails to learn it correctly. What is the probability that he was taught by method A?

Answer 1

Answer:

P(A | F) = 81.81%

There is 81.81% probability that worker was taught by method A given that he failed to learn it correctly.

Step-by-step explanation:

The failure rate is 15% for A which means that

P(F | A) = 0.15

The failure rate is 5% for B which means that

P(F | B) = 0.05

Method B is more expensive and hence is used only 40% of the time which means that

P(B) = 0.40

Which means that A is used the other 60% of the time

P(A) = 0.60

A worker is taught the skill by one of the methods but fails to learn it correctly.

We are asked to find the the probability that he was taught by method A.

So that means we want to find out

P(A | F) = ?

We know that according to Baye's rule,

$P(A \:|\: F) = \frac{P(F \:|\: A)\times P(A) }{P(F \:|\: A)\times P(A) + P(F \:|\: B)\times P(B) }$

Substitute the given probabilities into the above equation

$P(A \:|\: F) = \frac{0.15\times 0.60 }{0.15\times 0.60 + 0.05\times 0.40}\\\\P(A \:|\: F) = \frac{0.09 }{0.09 + 0.02}\\\\P(A \:|\: F) = 0.8181 \\\\P(A \:|\: F) = 81.81\%$

Therefore, there is 81.81% probability that worker was taught by method A given that he failed to learn it correctly.