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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
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 machineThis means that 3 cannot use the meat-cutting machine
We'll add this to our diagram:
Given: 18 can use the bread-slicing machineThis means that 2 cannot use the bread-slicing machine
We'll add this to our diagram:
Given: 15 can use both machinesWe'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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