Question 1205314: The table below shows the number of boxes of 100 tubes containing defectives:
Firm Number of defective tubes 0 1 2 3 or more
Supplier A 500 200 200 100
Supplier B 320 160 80 40
Supplier C 600 100 50 50
If one box had been selected randomly, what is the probability that
(a) the box would have come from Supplier A?
(b) it would have no defectives and come from Supplier A?
(c) it would have no defectives or come from Supplier A?
(d) it contained one or two defective tubes given the box came from Supplier B?
help me explain this thankyou
Answer by math_tutor2020(3817)   (Show Source):
You can put this solution on YOUR website!
Given table
| Number of defective tubes | | 0 | 1 | 2 | 3 or more | | Supplier A | 500 | 200 | 200 | 100 | | Supplier B | 320 | 160 | 80 | 40 | | Supplier C | 600 | 100 | 50 | 50 |
I recommend using a spreadsheet.
Let's compute the row and column totals.
| Number of defective tubes | | | 0 | 1 | 2 | 3 or more | Total | | Supplier A | 500 | 200 | 200 | 100 | 1000 | | Supplier B | 320 | 160 | 80 | 40 | 600 | | Supplier C | 600 | 100 | 50 | 50 | 800 | | Total | 1420 | 460 | 330 | 190 | 2400 |
The spreadsheet command SUM is very handy to add up a bunch of cells quickly.
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Part (a)
There are 1000 boxes from Supplier A.
This is out of 2400 boxes total.
1000/2400 = 5/12 is the probability of getting a box from Supplier A.
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Part (b)
There are 500 boxes that are from Supplier A and have no defective tubes.
This is out of 2400 boxes total.
500/2400 = 5/24
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Part (c)
W = set of boxes where all of the tubes are working perfectly
W = set of boxes that have no defects
n(W) = number of boxes that have no defects
n(W) = 1420 at the bottom of column 1 of the table.
A = set of boxes from Supplier A
n(A) = number of boxes from Supplier A
n(A) = 1000 at the end of row 1.
n(W and A) = 500 = number of boxes that have no defects and from Supplier A.
Use the inclusion-exclusion principle to compute the following.
n(W or A) = n(W) + n(A) - n(W and A)
n(W or A) = 1420 + 1000 - 500
n(W or A) = 1920
There are 1920 boxes that have all working tubes, come from Supplier A, or both.
Another approach could be to add up all of the values that are in the "Supplier A" row or "0 defects" column. Do not add in the row total or column total (since these totals are sums themselves).
500+200+200+100+320+600 = 1920
So this is another way to compute n(W or A) = 1920.
The first method is probably more standard in many textbooks.
After we find n(W or A) = 1920, divide this over the grand total 2400.
1920/2400 = 4/5
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Part (d)
We focus entirely on the "Supplier B" row. Use pieces of paper to cover up the other rows if they are a distraction.
There are 160+80 = 240 boxes that have either 1 defective tube or 2 defective tubes out of 600 boxes from this supplier.
240/600 = 2/5
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Answers:
(a) 5/12
(b) 5/24
(c) 4/5
(d) 2/5
I'll let the student convert these fractions to their decimal form if needed.
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