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?

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

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

Finally

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

d

step-by-step explanation:

given

y + 3 = - 4(x - 2) ← distribute the parenthesis

y + 3 = - 4x + 8 ( subtract 3 from both sides )

y = - 4x + 5 → d

ou would have 12 of each