Question 1189386: Hello, sorry for posting this again, just having a hard time understanding it so it will be really cool if someone can explain how they come up with the answer, thank you in advance.
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* Explain your answer after the solution
1. A shipment of 10 TV sets contains 3 defective units. If three units are taken for inspection, then what is the probability that:
a. all the defective TV sets are included
b. no defective TV set shall be included
c. only one of the defective TV sets shall be included
Answer by math_tutor2020(3817)   (Show Source):
You can put this solution on YOUR website!
I assume you are referring to this previous solution here
https://www.algebra.com/algebra/homework/Probability-and-statistics/Probability-and-statistics.faq.question.1189346.html
which admittedly is a bit cryptic. I'll try my best to decode what the tutor @Boreal is saying.
In case you're wondering, his/her steps and final answers are correct.
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Problem 1, part A
There are 10 TVs total and 3 of them don't work.
If we pick one TV purely at random, the probability of getting a defective TV is 3/10.
We simply list the number of ways to get what we want (the number defective) over the number total to form the probability needed.
As a slight tangent, it seems a bit weird to "want" a broken TV, but that's what part A calls for. Anyways, I digress.
After that first broken TV is selected, we have 3-1 = 2 broken TVs left in the lot out of 10-1 = 9 overall.
This leads directly to the fraction 2/9 to represent the probability of getting another broken TV.
All of this hinges on the fact that the first TV is not put back or replaced. Hence we use the idea of no replacement here.
Lastly, the third selection will only have one choice as there's only one broken TV left. This is out of 9-1 = 8 leftover.
We get the fraction 1/8 to represent the chances of getting yet another broken TV (three broken TVs in a row, talk about bad luck huh?)
To wrap things up, we have these fractions:To represent the probabilities of - getting the first TV broken
- AND getting the second TV broken
- AND getting the third TV broken
all in a row and in that order.
We then multiply out those fractions which is what the tutor @Boreal then wrote
(3/10)*(2/9)*(1/8) = (3*2*1)/(10*9*8) = 6/720 = 1/120
The nice thing here is that we have both the numerator and denominator count down by 1 each time we multiply a new item.
Answer: 1/120
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Problem 1, part B
There are 10 TVs total and 3 of them don't work. That means 10-3 = 7 of them do function properly.
If you are picking a single TV at random, the chances of getting a working TV is 7/10.
Once that working TV is picked, the fraction 6/9 represents the probability of getting a second working TV.
Like before, we have both the numerator and denominator count down by 1.
As expected, the fraction 5/8 is the probability of getting a third working TV in a row.
We have these fractionsto represent the probabilities of getting the first TV working, the second TV working, and the third TV working all in a row.
It might be tempting to reduce the fraction 6/9 to 2/3; however, I recommend keeping it unreduced so the fraction is more descriptive.
Like before, we'll multiply out those fractions
(7/10)*(6/9)*(5/8) = (7*6*5)/(10*9*8) = 210/720 = 7/24
Answer: 7/24
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Problem 1, part C
Let's say that the 1st slot is reserved for the broken TV, while the other two slots (2nd and 3rd) are for the working models.
Refer back to part A above to see that 3/10 is the probability of getting a broken TV.
Once that TV is picked, we have 7 working units out of 10-1 = 9 leftover. This gives the fraction 7/9.
We do not subtract 1 from 7 like we did with part B. At least not yet.
Once the second TV is picked, the probability of getting another working TV is 6/8 since we have 7-1 = 6 working TVs left out of 9-1 = 8 total. The denominators steadily decrease (10,9,8) but the numerators are a bit more nuanced. The general theme though is we have this countdown going on.
So,- 3/10 = probability of getting the first selection broken
- 7/9 = probability of getting the second selection working
- 6/8 = probability of getting the third selection working
This leads to (3/10)*(7/9)*(6/8) = (3*7*6)/(10*9*8) = 126/720 = 7/40
But wait, that's not the full story.
That result of 7/40 is the probability of getting that exact lineup of 1st broken, 2nd working, 3rd working in that exact order.
However, it does not apply to a lineup like 1st working, 2nd broken, 3rd working or to 1st working, 2nd working, 3rd broken
To encode the possible lineups, we can have this
BWW
WBW
WWB
where B stands for "broken" and W stands for "working" respectively.
The order matters between the B and W's; but with just the W's the order doesn't matter.
As you can probably guess, we multiply that 7/40 by 3 to account for all three possible cases.
Then we can fully wrap this up
3*(7/40) = 21/40
Side note: When @Boreal mentions the notation 3C1, s/he is accounting for the three cases mentioned, i.e. it represents the number of ways to pick exactly one broken TV from a pool of 3.
The 7C2 accounts for the number of ways to pick 2 working TVs out of 7 total.
For more info, check out the nCr combination formula.
Answer: 21/40
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