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I am working through a stat book trying to freshen up on my math. Here is a problem that is posed...

A shipment of 10 televisions includes three that are defective. In how many wats can a hotel purchase four of these sets and receivee at least two of the defective sets?

My main issue w/ these type questions is that there is so much room for interpretation. If we consider just good and bad then i believe the answer is

${{4} \choose {1} } $ + ${{4} \choose {2} } $ + ${{4} \choose {3} } $ = 11.

Writing out the matrix seems to prove this out.

But I think the question is really asking about unique tvs. How many unique sets of serial number could i wind up w/ given at least 2 are bad.

My answer would be

${{10} \choose {4} } $ - ${{7} \choose {4} } $ - ${{7} \choose {3} } $ = 210 - 35 - 35 = 140

My thinking is that if you take all the combinations, and subtract out those that you want to disregard, you should have your answer. If i pick 4 good tvs, or 3 good tvs, i wont have 2 bad tvs.

My question is this - is what i did valid? I have looked at the different theorems in the text I am reviewing and I dont see anything that really supports this.

If not, could someone point me in the right direction?

Thanks in advance.

Greg

greg
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  • I guess the problem is when you are picking up 3 good tvs, the fourth one can be good tv also. So you cannot subtract all the ${{7} \choose {3} }$ because some of them you have already subtracted in ${{7} \choose {4} }$. So the last term you have subtracted should be for 3 good tvs and 1 defective tv i.e. ${{7} \choose {3} }{{3} \choose {1} }$. And you get the same answer 70. Cheers! – Babai Dec 16 '13 at 19:01
  • corrected fat finger. My solution equals 140, not 40. – greg Dec 16 '13 at 19:31

2 Answers2

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The answer is ${7\choose2}\cdot{3\choose2}+{7\choose3}\cdot{3\choose1}$

Babai
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  • There are only two options: (1) We choose two defective out of three and two o.k. from the rest of 7. (2) 3 defective out of 3 (which can be done in exactly one way) and 1 o.k. from the rest 7(which can be done in exactly 7 ways. – Babai Dec 16 '13 at 18:54
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    Typo: in the second factor of the second term. $1$ must be replaced by $3$ there. – drhab Dec 16 '13 at 18:59
  • Thanks. That makes a great deal of sense. Ill come back when I have more rep and upvote. – greg Dec 16 '13 at 19:21
  • This answers (correctly) another question: how many ways are there to receive at least $2$ non-defective sets. – drhab Dec 17 '13 at 09:42
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Discern two possibilities:

1) $2$ defectives are received.

Then $2$ are chosen from $7$ (the non-defectives) and $2$ from $3$ (the defectives)

This can be done on $\binom{7}{2}\binom{3}{2}=63$ ways.

2) $3$ defectives are received.

Then $1$ is chosen from $7$ (the non-defectives) and $3$ from $3$ (the defectives)

This can be done on $\binom{7}{1}\binom{3}{3}=7$ ways.

So there are $70$ ways in total.

drhab
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  • Thanks. that makes a great deal of sense. Ill come back when I have more rep and upvote. – greg Dec 16 '13 at 19:22