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kitchen crew

by lukaswelker » Thu Apr 17, 2014 9:02 am
hello guys

Probabilities are not my strong suite and Here goes a question that proves me right;

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 chosen 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

Does anybody have a quick easy answer?
Cheers
Lukas
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by theCodeToGMAT » Thu Apr 17, 2014 9:20 am
17 = use meat cutting machine
18 = bread slicing Machine
15 = both

(17-15) = 2 can use only Cutting Machine
(18-15) = 3 can use only Slicing Machine
15 can use both

S0, {0} since20 = (2) + (15) + (3)
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by Brent@GMATPrepNow » Thu Apr 17, 2014 10:56 am
lukaswelker 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 chosen at random, what is the probability that the member chosen will be someone who cannot use either machine.

A) 0
B) 1/10
C) 1/7
D) 1/4
E) 1/3
One approach is to use the Double Matrix Method. This technique can be used for most questions featuring a population in which each member has two characteristics associated with it.
Here, we have a population of kitchen staff, and the two characteristics are:
- can use the meat-cutting machine or cannot use the meat-cutting machine
- can use the bread-slicing machine or cannot use the bread-slicing machine

There are 20 people altogether, so we can set up our diagram as follows:
Image

Given: 17 can use the meat-cutting machine
This means that 3 cannot use the meat-cutting machine
We'll add this to our diagram:
Image

Given: 18 can use the bread-slicing machine
This means that 2 cannot use the bread-slicing machine
We'll add this to our diagram:
Image

Given: 15 can use both machines
We'll add this to our diagram:
Image

Since we know the SUMS of the boxes in each column and in each row, we can complete the diagram as follows:
Image

As we can see, there are 0 people in the box representing someone who cannot use either machine.

So, P( selected person cannot use either machine) = [spoiler]0/20 = 0 = A[/spoiler]


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Difficult Data Sufficiency questions
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Brent
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by [email protected] » Fri Apr 18, 2014 1:02 am
Hi lukaswelker,

This question is actually an overlapping sets question. It can be solved in a couple of different ways. In addition to the tic-tac-toe board that Brent describes, you can also use an algebra equation:

Total = Group1 + Group2 - Both + Neither

Here, the...

Total = 20 members
Group 1 = meat-cutters = 17
Group 2 = bread-slicers = 18
Both = 15

20 = 17 + 18 - 15 + (people who use neither)

20 = 20 + Neither

Neither = 0

The probability of selecting someone who cannot use either machine is 0/20 = 0

Final Answer: A

As an aside, it would be unlikely for the GMAT to include a group of people with 0 members in it, so this question is not quite test-like.

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Rich
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