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Mamont248 [21]
2 years ago
9

Probability, help, please ​

Mathematics
1 answer:
Butoxors [25]2 years ago
5 0

Problem 1

<h3>Answer: Choice A)  1/3</h3>

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

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

<h3>Answer:  B) 2/3</h3>

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

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

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