Question

1. Again we have two methods, A and B, available for teaching a certain industrial skill. There is an 80% 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 25% of the time (method A is used the other 75% of the time).

A worker learned the skill successfully. What is the probability that he was taught by method A?
2. Again we have two methods, A and B, available for teaching a certain industrial skill. There is an 80% 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 25% of the time (method A is used the other 75% of the time).
What does a correct probability tree for this problem look like?

Answer 1

The probability that a worker was taught by method A given that he learned the skill successfully is 0.72.

Explanation:

(1)

The information provided is:

A = if method A is used

B = if method B is used

S = successfully learning the skill

P (A) = 0.75

P (B) = 0.25

P (S|A) = 0.80

P (S|B) = 0.95

Compute the probability that a worker was taught by method A given that he learned the skill successfully as follows:

`P(A|S)=(P(S|A)P(A))/(P(S|A)P(A)+P(S|B)P(B))`

`=(0.80* 0.75)/(0.80* 0.75+0.95* 0.25)\\\\=(0.60)/(0.60+0.2375)\\\\=(0.60)/(0.8375)\\\\=0.716418\\\\\approx 0.72`

Thus, the probability that a worker was taught by method A given that he learned the skill successfully is 0.72.

(2)

The probability is attached below.