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Math  /  Discrete

QuestionTime left 0:10:28
In a production line, 22 filling machines are used for cans filling. Four cans are selected at random and without replacement. Suppose that Five of the machines cause overweight that exceeds the customer requirements. In addition, three different machines cause underweight filling. How many possible samples such that exactly one can in the sample is considered overweight and exactly one can in the sample is considered underweight?

Studdy Solution

STEP 1

What is this asking? Out of 22 machines, 5 make cans too heavy and 3 make them too light.
If we grab 4 cans at random, what are the chances exactly one is too heavy and exactly one is too light? Watch out! Don't forget that we're picking cans without replacement, so the number of choices changes each time!
Also, we need *exactly* one overweight and one underweight can, not *at least* one.

STEP 2

1. Overweight Cans
2. Underweight Cans
3. Okay Cans
4. Total Combinations

STEP 3

Let's start by figuring out how many ways we can pick *exactly* one overweight can out of the 5\text{5} overweight machines.
Since we're picking only 1\text{1} can from the overweight group, we have (51)=5!1!⋅(5−1)!=5!1!⋅4!=5⋅4⋅3⋅2⋅1(1)⋅(4⋅3⋅2⋅1)=51=5 {5 \choose 1} = \frac{5!}{1! \cdot (5-1)!} = \frac{5!}{1! \cdot 4!} = \frac{5 \cdot 4 \cdot 3 \cdot 2 \cdot 1}{(1) \cdot (4 \cdot 3 \cdot 2 \cdot 1)} = \frac{5}{1} = \textbf{5} ways to choose one overweight can.

STEP 4

Now, let's do the same thing for the underweight cans.
We have 3\text{3} machines that produce underweight cans, and we want *exactly* one of them in our sample.
So, we have (31)=3!1!⋅(3−1)!=3!1!⋅2!=3⋅2⋅1(1)⋅(2⋅1)=31=3 {3 \choose 1} = \frac{3!}{1! \cdot (3-1)!} = \frac{3!}{1! \cdot 2!} = \frac{3 \cdot 2 \cdot 1}{(1) \cdot (2 \cdot 1)} = \frac{3}{1} = \textbf{3} ways to pick one underweight can.

STEP 5

We've picked one overweight can and one underweight can.
We need to pick 2\text{2} more cans to complete our sample of 4\text{4} cans.
Since we already have one overweight and one underweight can, these 2\text{2} cans must be from the "okay" machines.
There are 2222 total machines, 55 overweight, and 33 underweight, leaving 22−5−3=1422 - 5 - 3 = \textbf{14} "okay" machines.

STEP 6

We need to pick 2\text{2} cans from these 14\text{14} "okay" machines.
The number of ways to do this is (142)=14!2!⋅(14−2)!=14!2!⋅12!=14⋅132⋅1=7⋅13=91 {14 \choose 2} = \frac{14!}{2! \cdot (14-2)!} = \frac{14!}{2! \cdot 12!} = \frac{14 \cdot 13}{2 \cdot 1} = 7 \cdot 13 = \textbf{91} ways.

STEP 7

To get the total number of ways to pick exactly one overweight, exactly one underweight, and two "okay" cans, we multiply the number of ways to choose each type of can together: 5\textbf{5} (overweight) ⋅\cdot 3\textbf{3} (underweight) ⋅\cdot 91\textbf{91} (okay) =15⋅91=1365= 15 \cdot 91 = \textbf{1365} ways.

STEP 8

There are 1365\textbf{1365} possible samples with exactly one overweight can and exactly one underweight can.

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