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Math Help - I tried this question, but told I am close BUT not the correct answer. Help please.

  1. #1
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    Exclamation I tried this question, but told I am close BUT not the correct answer. Help please.

    question goes like this.

    Food inspector examines 3 products in a group to determine if they are acceptable. From past experience it is known that the probability of finding no defects in the sample of 3 if 0.990, the probability of 1 defect in the sample of 3 is 0.006, and the probability of finding 2 defects in the sample of 3 is 0.003.

    What is the probability of finding 3 defecting unites in the sample of 3??

    I did .99 x .006 x .003 = .00001782 CLOSE but not the correct answer. I am not getting my stats class, something is not clicking. There is a piece of information that is not helping me connect all of this.

    So what am I doing wrong?Or is it 3 times each number .99, .006, .003?

    Any help would be great, thank you.
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  2. #2
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    Re: I tried this question, but told I am close BUT not the correct answer. Help pleas

    Isn't it true that 0 defects, (exactly) 1 defect, 2 defects and 3 defects are four disjoint events that cover the whole sample space?
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  3. #3
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    Re: I tried this question, but told I am close BUT not the correct answer. Help pleas

    Can you explain what you wrote? 3 non conformning units is asking DEFECTIVE correct? in a sample of 3.

    Funny how I can do thermodynamics, but can't do stats. GO FIGURE (scratching head).
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    Re: I tried this question, but told I am close BUT not the correct answer. Help pleas

    What emakarov wrote basically means that there are four, and only four possible outcomes for the 3 units, and that they don't overlap in any way (i.e. it's the probability of having *exactly* 1 defect, rather than *at least* 1 defect). You're told the probability of finding no defects, 1 defect, and 2 defects; the only remaining possibility is 3 defects (as there are only 3 units in the sample). Now you just need to recall that the probability of all possible events must add up to 1 (i.e P(none) + P(one) + P(two) + P(three) = 1). At this point you have all the information you need to determine P(three).
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    Re: I tried this question, but told I am close BUT not the correct answer. Help pleas

    Quote Originally Posted by bradycat View Post
    There are only 0,1,2 defects FYI.
    If I understand correctly, 3 defects are also possible.

    What I mean, briefly, is that one cannot have exactly 2 and exactly 3 defects (and similarly for other non-equal numbers). Also, the possible number of defects is from 0 to 3; there are no other options. So, the probabilities for different number of defects should add up to 1.

    More formally, we can order the three products and say that an examination produces an ordered triple (r_1, r_2, r_3) where r_i=1 if product #i is defect-free and r_i=0 otherwise. Then the sample space (the set of all possible outcomes) is \Omega=\{0,1\}^3, i.e., the set of all triples of 0 and 1. The event "a sample of 3 products has exactly n defects" is the subset of Ω that includes triples having exactly n 1's. Though there may be a different construction of the sample space here, the main point is that those events (subsets) for n = 0, 1, 2, 3 are disjoint and their union is the whole Ω. Therefore, their probabilities add up to 1.
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    Re: I tried this question, but told I am close BUT not the correct answer. Help pleas

    SO the answer to this is.

    p(.99)+p(.006)+p(.003)+p(3 defects) = 1
    So need to solve for P which is .001??

    Question, is there a time when we use addition, then multiplication? In the examples the teacher said it was close to used multiplication, why I used it above. But reading in my book to use that is when they are independent and do not affect and in sequence?
    So they have to be two different things that have no association to each other?? like a dice and card, where if you have a card and another card, not the same?

    Just trying to understand, any help would be good. I have even gone on websites, but they are just basics, nothing in depth to what I am looking for.

    Again, thanks.
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  7. #7
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    Re: I tried this question, but told I am close BUT not the correct answer. Help pleas

    Britmath...... so how would it be worded to know it was multiplication? maybe that will help me decipher it a bit more.
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  8. #8
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    Re: I tried this question, but told I am close BUT not the correct answer. Help pleas

    Quote Originally Posted by bradycat View Post
    SO the answer to this is.

    p(.99)+p(.006)+p(.003)+p(3 defects) = 1
    So need to solve for P which is .001??
    You're correct that the answer is 0.001, though I should note that you are not solving for P in this instance; the notation P(one), etc is just a short hand for "the probability of one defect"), so P(none) + P(one) + P(two) + P(three) = 1 can be restated as simply 0.99 + 0.006 + 0.003 + x = 1, solve for x.

    As for the issue of addition and multiplication, and dependent and independent events, I'll try to give some examples to clarify.

    The use of addition and multiplication depends on whether the events are alternatives to each other, or whether they can occur together.

    Consider a roll of a fair six-sided die. Note that the events are independent, in that the result of each throw has no impact on the probability of the throws that follow it. The probability of getting a 1 (or any other number for that matter) is obviously \frac{1}{6}.
    If you are then asked to determine the probability of throwing 3 consecutive 1s, that is 1 AND 1 AND 1, then the probability is \frac{1}{6}*\frac{1}{6}*\frac{1}{6}&=\frac{1}{216}.

    Another case might be to determine the probability that in a single throw of the die you obtain a 1 OR 3. In this instance you take the probability of each event and add them \frac{1}{6}+\frac{1}{6}&=\frac{2}{6}&=\frac{1}{3}.

    A case involving both addition and multiplication might be the following: Find the probability of obtain a 1 OR 3 on the first throw, AND a 5 on the second throw. Here, the probability becomes \left(\frac{1}{6}+\frac{1}{6}\right)*\frac{1}{6}&=  \frac{2}{36}&=\frac{1}{18}. Here you sum the probabilities for the OR part, and then multiply the result of that by the probability of throwing a 5, because it's (1 OR 3) AND 5. I hope that's helpful for independent events. I'll put dependent events in a separate post.
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    Re: I tried this question, but told I am close BUT not the correct answer. Help pleas

    Since I have already typed it...

    Quote Originally Posted by bradycat View Post
    p(.99)+p(.006)+p(.003)+p(3 defects) = 1
    So need to solve for P which is .001??
    Usually p is a function that maps events (subsets of the sample space) to numbers, so p(3 defects) may be OK, but p(.99) does not make sense. Yes, the answer is .001.

    Quote Originally Posted by bradycat View Post
    Question, is there a time when we use addition, then multiplication?
    If A, B and C are events such that AB (the intersection of A and B) and C are disjoint and A and B are independent, then P(AB or C) = P(A)P(B) + P(C).

    Quote Originally Posted by bradycat View Post
    But reading in my book to use that is when they are independent and do not affect and in sequence?
    Yes, the probability of independent events is the product of probabilities. Independent events may occur in sequence or simultaneously. For example, if you toss two coins simultaneously, the events that the coins land head up are independent.

    Quote Originally Posted by bradycat View Post
    So they have to be two different things that have no association to each other?? like a dice and card, where if you have a card and another card, not the same?
    Mathematically, events A and B are independent, by definition, if P(AB) = P(A)P(B). But yes, in solving problems, we usually say that events are independent not by checking their probabilities, but from physical considerations, if one event cannot influence the other.

    Quote Originally Posted by bradycat View Post
    Just trying to understand, any help would be good. I have even gone on websites, but they are just basics, nothing in depth to what I am looking for.
    There is no substitute for reading a good source, like a textbook or lecture notes. This site is only for "marginal legal advice."
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  10. #10
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    Re: I tried this question, but told I am close BUT not the correct answer. Help pleas

    As for dependent events, you're now considering scenarios where the outcome of one event impacts the probability of events that follow it. A classic case is a bag containing Red and Blue balls, which are removed, *without* replacement. The use of multiplication and addition is no different than for independent events (i.e. AND uses multiplication, OR uses addition), the difference is that the actual probabilities of events will not remain the same.

    As an example, consider a bag containing 17 balls; 10 Red and 7 Blue. Before any balls are removed from the bag, the probability that a ball, selected at random, will be Red is \frac{10}{17}, while the probability that it will be Blue is \frac{7}{17}.

    A ball is now removed, without replacement, which happens to be Red, obviously there are now only 16 balls in the bag, 9 now being Red, and still 7 Blue, so the probabilities become:

    P(Red)&=\frac{9}{16}\\P(Blue)&=\frac{7}{16}

    So, if we now consider the bag in its original state and ask what is the probability that a Red, followed by a Blue, followed by another Blue will be drawn at random from the bag, without replacement? Hopefully it's fairly clear that the problem looks like this:

    \frac{10}{17}*\frac{7}{16}*\frac{6}{15}&=*\frac{42  0}{4080}&=\frac{7}{68}

    I hope that's clear (incidentally, I agree with emakarov; I'm happy to help with improving conceptual understanding, but I'm just a guy with an Internet connection , so caveat emptor and all that!).
    Last edited by britmath; May 31st 2012 at 01:40 PM.
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  11. #11
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    Re: I tried this question, but told I am close BUT not the correct answer. Help pleas

    I will go thru this, THANK you for your help from you both. This is good!!!
    Last edited by bradycat; May 31st 2012 at 02:38 PM.
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  12. #12
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    Re: I tried this question, but told I am close BUT not the correct answer. Help pleas

    THANK YOU AGAIN, just read thru it and I understand it a heck of a lot better now.

    Appreciate the time you took in replying.
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