Question

Two methods, A and B, are available for teaching a certain industrial skill. The failure rate is 35\% for A and 15\% for B. However, B is more expensive There hence is used only 40\% of the time, while A is used 60\% of the time. 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:

Therefore the probability that he was taught by method A is 0.78.

Explanation:

Probability:

The ratio of the number of favorable outcomes to the number of all possible outcomes.

Bayes' Rule:

If the events $B_1$,$B_2$, .....$B_n$ from a portion of a sample space S and A is any events of A, then

$P(B_i|A)=\frac{P(B_i)P(A|B_i)}{\sum_{j=1}^kP(B_j)P(A|B_j)}$

Given that,

There are two available method for teaching A and B.

The failure rate for A is 35%

That is P(F|A) =35%=0.35

The failure rate for B is 15%

That is P(F|A) =15%=0.15

A used 40% of the time.

P(A)=40%=0.40

B used 60% of the time.

P(A)=60%=0.60

To find P(A|F) , we use the Bayes's rule.

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

            $=\frac{0.60\times 0.35}{0.60\times 0.35+0.40\times 0.15}$

           =0.78

Therefore the probability that he was taught by method A is 0.78.