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Math Help - At Least One - Probability Question

  1. #1
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    At Least One - Probability Question

    First of all I would like to say hello as I am a new member of this forum. I have a question that I am unable to obtain information on.

    I am in a Statistics class and after searching through my book and notes I still can not find the answer to this simple question. Hopefully someone can help me with this.

    Question: A box contains 10 items, 2 of which are defective. Two are drawn with replacement.

    The question is asking what is the probability of drawing at least one defective.

    I'm not looking for JUST an answer, but a formula perhaps that could help me understand and solve similar question of this sort.

    Any help would be greatly appreciated!
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  2. #2
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    This is WITH replacement, so this is a binomial.

    \sum_{k=1}^{2}C(2,k)(1/5)^{k}(4/5)^{2-k}

    C(2,1)(1/5)^{1}(4/5)^{1}+C(2,2)(1/5)^{2}(4/5)^{0}=\boxed{\frac{9}{25}}

    You could also find the probability of none and subtract from 1. That's the easiest. Anytime you see 'at least one'. Find the prob. of none and subtract from one. That is usually a better way to go.

    1-C(2,0)(1/5)^{0}(4/5)^{2}=\frac{9}{25}
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  3. #3
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    Quote Originally Posted by galactus View Post
    This is WITH replacement, so this is a binomial.

    \sum_{k=1}^{2}C(2,k)(1/5)^{k}(4/5)^{2-k}

    C(2,1)(1/5)^{1}(4/5)^{1}+C(2,2)(1/5)^{2}(4/5)^{0}=\boxed{\frac{9}{25}}

    You could also find the probability of none and subtract from 1. That's the easiest. Anytime you see 'at least one'. Find the prob. of none and subtract from one. That is usually a better way to go.

    1-C(2,0)(1/5)^{0}(4/5)^{2}=\frac{9}{25}

    Thanks for the quick response, I think I am a little more comfortable using the second method, but could you break it down for me and explain the numbers that you plugged in, because I am still a little confused.
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  4. #4
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    It's just the binomial probability formula. It'll be in your stats book. The 1/5 is the prob. of getting a defect, p = 2/10 = 1/5.; q = 1-p = 1-1/5 = 4/5
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  5. #5
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    Thanks for your help, I will check it out!

    Let me bother you with one more similar question, but slightly different. This one is WITHOUT replacement.

    "A bin contains 5 defective and 15 non-defective batteries. If two batteries are selected at random without replacing the first, what is the probability?"

    At least one is defective...
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  6. #6
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    This requires a hypergeometric probability. It's like a binomial, but without replacement.

    Again, the best bet is to find prob. of none and subtract from 1.

    P(x)=\frac{(C(k,x))(C(N-k,n-x))}{C(N,n)}

    This will also be in your stats book.

    N=population
    n=sample size
    k=successes

    Given a population of N items having k successes and N-k failures, you must
    Find the prob of selecting n items from N that has x successes and n-x failures.

    P(at least one defect)=
    1-\left[\frac{C(5,0)C(15,2)}{C(20,2)}\right]
    Last edited by galactus; October 22nd 2007 at 10:28 AM.
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  7. #7
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    Hello, edouble2k6!

    A bin contains 5 defective and 15 non-defective batteries.
    If two batteries are selected at random without replacing the first,
    what is the probability that at least one is defective?

    There are: .5 bad and 15 good batteries.

    There are: . {20\choose2} = 190 possible two-bettery samples.


    The opposite of "at least one bad" is "no bad" or "both good".

    There are: . {15\choose2} \:=\:105 ways to draw two good batteries.

    Hence: . P(\text{0 bad}) \:=\:\frac{105}{190} \:=\:\frac{21}{38}

    Therefore: . P(\text{at least one bad}) \:=\;1 - \frac{21}{38} \:=\:\frac{17}{38}

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