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sabineodf
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Of the 20 members of a kitchen crew, 17 can use the meat-cutting machi 24 May 2015, 19:39
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88% (01:02) correct 12% (01:26) wrong based on 678 sessions

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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 is 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
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A
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Of the 20 members of a kitchen crew, 17 can use the meat-cutting machi 22 Sep 2016, 13:27
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sabineodf
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 is 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
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Here's a step-by-step approach using the Double Matrix method.

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:


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:


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:


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


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


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) = 0/20 = 0

Answer: A

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Of the 20 members of a kitchen crew, 17 can use the meat-cutting machi 24 May 2015, 19:59
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sabineodf
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 is 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
Show more

You know how to solve for Total number of people who can use at least one machine:

Total = n(A) + n(B) - n(A and B)

Total people who can use at least one machine = 17 + 18 - 15 = 20

So all 20 people know how to use at least one machine. Hence, if you pick anyone, the probability is 0 that he wouldn't know how to use any machine.

Answer (A)
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Of the 20 members of a kitchen crew, 17 can use the meat-cutting machi 24 May 2015, 21:27
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Hi sabineodf,

This question is modeled after the typical Overlapping Sets prompts that the GMAT includes on Test Day (and you'll see at least one of those on the Exam). The numbers in the prompt are relatively small though, so you can don't actually need any special organization to get to the correct answer - you just need some basic logic and note-taking:

We know there are 20 total people and that 15 of those people can use BOTH machines.....

That leaves 5 people....

Since 17 can use the meat-cutting machine.....that means that 2 of those people CAN'T use both machines.
Since 18 can use the bread-cutting machine....that means that 3 of those people CAN'T use both machines.

The group can thus be described as....
15 can use BOTH machines
2 can use JUST the meat-cutting machine
3 can use JUST the bread-cutting machine

15+2+3 = 20......so ALL 20 people have been accounted for and 0 people can use NEITHER machine. Thus, the probability of selecting someone who cannot use either machine is 0%.

Final Answer:
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A

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Of the 20 members of a kitchen crew, 17 can use the meat-cutting machi 17 Sep 2022, 03:52
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Hi,

How do we know that "either" means BOTH and not at least 1 of the two machines? If at least one, the answer would be 1/4.
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