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In this problem, it is said that a company has two machines that produce widgets.
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We are given information about how many defective widgets each machine produces, and we need to find the probability that a randomly chosen widget produced by the company is defective.
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Now, let us consider a few events.
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Let o be the event that the widget is produced by the old machine.
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Let n be the event that the widget is produced by the new machine.
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And let d be the event that the widget is defective.
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Now, note that o and n, these two events are mutually exclusive and exhaustive events.
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They are mutually exclusive because they cannot happen together because one widget will only be produced by one machine.
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And they are exhaustive because we only have these two machines.
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There are no other machines, so at least one of these two events must occur.
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Now, since o and n are mutually exclusive and exhaustive, if we add their probabilities, we will get one.
01:03
And this is because o and n being mutually exclusive and exhaustive will form a partition of the sample space.
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So here it is said that the new machine produces four times as many widgets as the older machine does.
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So the probability that a widget is produced by the new machine will be four times the probability that it is produced by the older machine.
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So that means we have p of o and p of n will be four times p of o.
01:32
So this is what we have.
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So on the left hand side we'll end up with five times p of o and this is equal to 1.
01:39
And this means that the probability of o is equal to 1 divided by 5 and that's equal to 0.
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0 .2.
01:49
And p of n is 4 times of p of o...