Statistics with Probability

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  • #12867

    Alphatron
    Participant

    I’m having trouble understanding how to go about solving this problem. If I was simply looking for the number of combinations that could be made by choosing 2 items out of the 120, then the answer would be (120 * 119)/(2*1), correct? But how do I figure out the odds of picking one of the 10 out of the 120?

    #15905

    Alphatron
    Participant

    Wait? Maybe I’ve been going about this the wrong way? Am I just supposed to multiply the chances? So on the chance that I don’t pick a single defective item, it would be (11/12) * (109/119)?

    What do you do in the instance that the first item isn’t defective, but the second one is? Whatever I try, when I add the possibilities of both items not being defective, both items, being defective, and either being defective, I don’t get 1.0.

    #15906

    dannyCR09
    Member

    I think the way you solve this problem is that you have to look for the chance of getting a certain number of chips that are defective. Since X can only be 0, 1, or 2: you have the find the chance of either being defective. So for finding the probability of finding 1 item being defective, you would multiply the combination(10,1) by 110,1) since there are 10 defective chips and 110 good chips and then divide by the combination(120,2) since you are only testing 2 as a sample out of the 120 chips total.

    #15909

    Ezra Halleck
    Participant

    In your second post alphatron, I think that I can bring up the analogy of flipping a coin twice. The outcomes are HH, HT, TH and TT. So for our problem, the outcomes are DD, DG, GD, GG where D is defective and G is good. The one thing to keep in mind, is that unlike a coin toss problem, the probabilities change depending on what happened for your first choice. However, they should still not be that difficult to figure out. Once you have figured out the probabilities then you combine the DG and GD into a single category, namely 1 defective.

    Danny is basically providing you with a much easier approach, which is to use something called a hypergeometric distribution. We will not cover it formally in class, but you might look online for a discussion. There is a section in the textbook but I would not recommend it. Basically, you have fraction of a product of 2 binomial coefficients over a third binomial coefficient, namely the one you allude to in your first post. In the numerator, you select a certain number of defective mutliplied by a selection of a certain number of good ones: combin(total number of defective, number of defective you are selecting)*combin(total number of good, number of good you are selecting). The denominator is a blind selection: combin(total number of objects, size of selection without dividing into good and defective). How many good and defective you select depends on what your focus is (on defective) and which outcome of the random variable you are looking at (0,1,2). For example, if the outcome is 1, then you are selecting one defective and one good. If you the outcome is 2, then you are selecting 2 defective and 0 good.

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