Answered

At Westonci.ca, we make it easy to get the answers you need from a community of informed and experienced contributors. Our platform provides a seamless experience for finding precise answers from a network of experienced professionals. Discover in-depth answers to your questions from a wide network of professionals on our user-friendly Q&A platform.

Probability, help, please

​

Probability Help Please class=

Sagot :

Problem 1

Answer: Choice A)  1/3

-------------------

Explanation:

The notation "P(tune-ups | AC repair)" is the same as saying "P(tune-ups given AC repair)".

The "given" means we only focus on the "AC repair" circle.

Add up the numbers in that circle to get: 1+3+2+0 = 6

There are 6 ASE certified mechanics that can do AC repair.

Of those 6 people, only 2+0 = 2 can do engine tune-ups. Note I added the values in the "AC repair" and "tune-ups" overlapped region.

Therefore, we get the probability 2/6 = 1/3

==========================================================

Problem 2

Answer:  B) 2/3

-------------------

Explanation:

First, we'll add the numbers found in either the "brakes" or "tune-ups" circles.

4+1+2+2+3+0 = 12

There are 12 mechanics that can handle brakes or tune-ups or both.

Now add up all of the numbers shown regardless of their location. We have 4+1+3+2+2+0+3+3 = 18 mechanics total.

So,

m = number who can do brakes or tune-ups or both

m = 12

n = number of mechanics total

n = 18

And,

P(brakes or tune-ups) = m/n

P(brakes or tune-ups) = 12/18

P(brakes or tune-ups) = (6*2)/(6*3)

P(brakes or tune-ups) = 2/3

Thank you for trusting us with your questions. We're here to help you find accurate answers quickly and efficiently. Thank you for choosing our platform. We're dedicated to providing the best answers for all your questions. Visit us again. Get the answers you need at Westonci.ca. Stay informed by returning for our latest expert advice.