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

Answer:

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

Step-by-step 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:

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

            $=\frac{0.80\times 0.75}{0.80\times 0.75+0.95\times 0.25}\\\\=\frac{0.60}{0.60+0.2375}\\\\=\frac{0.60}{0.8375}\\\\=0.716418\\\\\approx 0.72$

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.