Question

We have two methods A and B available for teaching a certain industrial skill. There is a 75% chance of successfully learning the skill if method A is used, and a 95% chance of success if method B is used. However, method B is substantially more expensive and is therefore used only 20% of the time(method A is used the other 80% of the time) A worker learned the skill successfully.
What is the probability that he was taught by method A?

Answer 1

Answer:

75.95% probability that he was taught by method A.

Step-by-step explanation:

We have these following probabilities:

An 80% probability of method A being used.

A 20% probability of method B being used.

If method A is used, a 75% probability of learning the skill successfully.

If method B is used, a 95% probability of learning the skill successfully.

This exercise can be formulated as the following problem:

What is the probability of B happening, knowing that A has happened.

It can be calculated by the following formula

$P = \frac{P(B).P(A/B)}{P(A)}$

Where P(B) is the probability of B happening, P(A/B) is the probability of A happening knowing that B happened and P(A) is the probability of A happening.

So

What is the probability of him being taught by method A, given that he learned the skill succesfully.

P(B) is the probability of being taught by method A, so P(B) = 0.80.

P(A/B) is the probability of learning the skill successfully, given that he was taught by method A. So P(A/B) = 0.75.

P(A) is the probability of learning the skill successfully. It is 75% of 80% plus 95% of 20%. So

$P(A) = 0.75*0.8 + 0.95*0.2 = 0.79$

Finally

$P = \frac{P(B).P(A/B)}{P(A)} = \frac{0.8*0.75}{0.79} = 0.7595$

There is a 75.95% probability that he was taught by method A.