Probability Problem Using Formula for Combinations

In summary: And thanks for the compliment, but I'm sure your teacher is doing a fine job. :) In summary, the problem is to find the probability of selecting one defective compressor out of a sample of 6 refrigerators, when there are 5 defective compressors out of a total of 12 refrigerators. This can be done by finding [(5 choose 1)(7 choose 5)]/(12 choose 6), which yields approximately 0.1136. To find the expected value of X, the number of defective compressors in a sample of 6, we use the formula E(X) = Σ[x * p(x)]. This involves finding p(1) through p(5), and multiplying each by the corresponding value
  • #1
Shoney45
68
0

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.
 
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  • #2
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
Dick said:
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
Dick said:
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
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
Shoney45 said:
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
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
Shoney45 said:
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
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
Shoney45 said:
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.
 

What is the formula for combinations?

The formula for combinations is nCr = n! / r!(n-r)!, where n represents the total number of objects and r represents the number of objects being chosen.

How do you apply the formula for combinations to a probability problem?

To apply the formula, you must first determine the total number of objects and the number of objects being chosen. Then, plug those values into the formula to calculate the number of possible combinations. Finally, divide the number of desired outcomes by the total number of possible combinations to find the probability.

Can the formula for combinations be used for both ordered and unordered selections?

No, the formula for combinations is only used for unordered selections. For ordered selections, the formula for permutations should be used instead.

What is the difference between combinations and permutations?

Combinations are unordered selections of objects, while permutations are ordered selections. The formula for combinations does not take into account the order in which the objects are chosen, while the formula for permutations does.

Can the formula for combinations be used for any type of problem?

No, the formula for combinations can only be used for problems where the order of the objects being chosen is not important. If order does matter, the formula for permutations should be used instead.

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