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Probability Problem Using Formula for Combinations

  • Thread starter Shoney45
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Homework Statement

Each of 12 refrigerators of a certain type has been returned to a distributor because of the presence of a high pitched oscillating noise. Suppose that 5 of these 12 have defective compressors and the other 7 have less serious problems. If they are examined in random order, let X = the number among the first 6 examined that have a defective compressor. Compute the following:

P(x = 1)

E(X)

E(X^2)

Homework Equations





The Attempt at a Solution

I'm getting really hung up here. I know my sample size is 6 refrigerators. 5 of the twelve are defective. My chances of picking a defective unit on the first try is 5/12.

Someone told me that it was [(5 choose 3)(7 choose 3)]/(12 choose 6). I just don't understand how to do this well enough, so I don't know if it is the right answer or not. And I want to know how to do it anyway.
 

Answers and Replies

  • #2
Dick
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Don't double post the same question. Ok? So what is [(5 choose 3)(7 choose 3)]/(12 choose 6) supposed be the answer to? Can you delete any other posts you have on this question and just concentrate on one?
 
  • #3
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Don't double post the same question. Ok? So what is [(5 choose 3)(7 choose 3)]/(12 choose 6) supposed be the answer to? Can you delete any other posts you have on this question and just concentrate on one?
To get one bad compressor in a sample of 6, you have to choose 1 from the group of 5 that have bad compressors and 5 from the group of 7 that don't. Then to get the probability you divide by the number of ways to choose 6 from the 12 sample total. How do you express that in combinations language?

My apologies. I didn't even realize I had posted my question once before. I put your answer to that one here. So I'll go and delete that one. I'm sorry about that. One of those days.
 
  • #4
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Don't double post the same question...Can you delete any other posts you have on this question and just concentrate on one?
I'm working the hint you gave me in my first (forgotten) post. While I work through it, I would like to delete my old post, but can't figure out how to do it. I can edit it, but I can't figure out how to delete it.
 
  • #5
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Okay, so I set up the problem as [(5 choose 1)(7 choose 5)]/(12 choose 6) which yields appx 0.1136.

Now I have to find E(X) for this. I know that E(X) = The Summation of [x*p(x)]. I don't know what x, or p(x) is though. Am I supposed to be using the probability that I found from this prior question?
 
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  • #6
Dick
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Okay, so I set up the problem as [(5 choose 1)(7 choose 5)]/(12 choose 6) which yields appx 0.1136.

Now I have to find E(X) for this. I know that E(X) = The Summation of [x*p(x)]. I don't know what x, or p(x) is though. Am I supposed to be using the probability that I found from this prior question?
So far you've just found p(1). How many more do you need to find to compute E(X)? What's the range of possibilities of the number of bad compressors in a sample of 6?
 
  • #7
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So would I need to find p(2) through p(6) [p(6) would be zero, so it is really just through p(5)].

Then once I do this. I need to multiply x(subscript i) * p(x(subscript i).

If I'm following this correctly, then I will have (1*.1136) + [2*p(x2)] + ...+ [6*p(x6)].

Am I on the right track here?
 
  • #8
Dick
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So would I need to find p(2) through p(6) [p(6) would be zero, so it is really just through p(5)].

Then once I do this. I need to multiply x(subscript i) * p(x(subscript i).

If I'm following this correctly, then I will have (1*.1136) + [2*p(x2)] + ...+ [6*p(x6)].

Am I on the right track here?
Sure, just like your Poisson problem, it's not that hard once you know what the parts are. Don't forget there is a p(0) too (which doesn't contribute to E(X)).
 
  • #9
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Beautiful! So after I have done this step, then I find E(X^2) by doing the exact same thing, except that the value of X is squared right?

I wish you were my teacher instead of this other guy.
 
  • #10
Dick
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Beautiful! So after I have done this step, then I find E(X^2) by doing the exact same thing, except that the value of X is squared right?

I wish you were my teacher instead of this other guy.
Yes, that's what you do.
 

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