Problem Solving

Of the 20 members of a kitchen crew, 17 can use the meat-cutting machine, 18 can use the bread-slicing machine, and 15 can use both machines. If one member of the crew is to be chose at random, what is the probability that the member chosen will be someone who cannot use either machine?

• 0
  1/10
  1/7
  1/4
  1/3

This question is from GMATPrep's Exam Pack 3 and I cannot solve it for the life of me. Please provide an explanation. Thanks!

Answer is A

Use a Matrix Method..
Black ones are from question
Green ones are implied
Red Box is what we need to find.

[spoiler]{A}[/spoiler]

I agree, you need to take your time on this or it's very easy to make a mistake.

Think of it like a box where you are adding up numbers to get to 20.

So you know the people who can use both machines is 15 - this tells us that there is an overlap of 15 for both sides.

[ 15 ]

Then, we look at meat cutters (17), so the excess would be 2

[ 15 ]+[ 2 Meat cutters]

and then the difference between both and Bread makers which equals 3

[ 15 both ]+[ 2MC ]+[ 3BM ] = [ 20 total ]

Since the total is 20, there is no way that you could not pick someone that is not qualified in either. Hope this helps!

chiphipoke wrote:Of the 20 members of a kitchen crew, 17 can use the meat-cutting machine, 18 can use the bread-slicing machine, and 15 can use both machines. If one member of the crew is to be chose at random, what is the probability that the member chosen will be someone who cannot use either machine?

• 0
  1/10
  1/7
  1/4
  1/3

We can use the following formula for two overlapping groups:

Total = Group 1 + Group 2 - Both + Neither.

The big idea with overlapping groups is to SUBTRACT THE OVERLAP.
When we count everyone in Group 1 (meat-cutters) and everyone in Group 2 (bread-slicers), those in BOTH groups (members who can use BOTH types of machines) get counted twice.
So that we don't double-count the members in both groups, we SUBTRACT THE OVERLAP from the total.

In the problem at hand:
Total = 20.
Group 1 = meat-cutters = 17.
Group 2 = bread-slicers = 18.
Both = 15.
Let N = the number who can use neither type of machine.

Plugging these values into the equation above, we get:
20 = 17 + 18 - 15 + N
N = 0.

Since 0 members can use neither type of machine, P(selecting a member who can use neither type of machine) = 0.

The correct answer is A.

One last tip: even if you're completely lost on test day, you KNOW the answer can't be (C) or (E) because you can't have (1/7) or (1/3) of a group of 20 people - no fractional persons allowed! You know that the answer can't be (1/4): since you have 18 people out of 20 who can use the bread slicer, you have AT MOST 2 people (or 1/10 of the total) who can't use either machine.

With that, you know the answer is either A or B, and you have a 50% chance of getting the question right! Pretty cool, and almost more satisfying than actually knowing how to solve.

chiphipoke wrote:Of the 20 members of a kitchen crew, 17 can use the meat-cutting machine, 18 can use the bread-slicing machine, and 15 can use both machines. If one member of the crew is to be chose at random, what is the probability that the member chosen will be someone who cannot use either machine?

• 0
  1/10
  1/7
  1/4
  1/3

This question is from GMATPrep's Exam Pack 3 and I cannot solve it for the life of me. Please provide an explanation. Thanks!

Answer is A

I was racking my brain on this problem, until I figured out you can use a Venn Diagram to make life easier!!

Had I not been so narrow-minded and always trying to solve problems using ONLY 1 method, I would have gotten this within seconds!

Petro, that's a great idea: any time you have two groups with some overlap, a Venn (or a matrix) is a lifesaver.