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Thread: Probability word problem

  1. #1
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    Probability word problem

    63% of people are happy. Pick 10 people randomly.
    a) What are the odds that 8 of them are happy?
    b) What are the odds that at least 2 of them are happy?
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  2. #2
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    Re: Probability word problem

    this is a binomial distribution with parameters $p=0.63,~n=10$

    can you solve it now?
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  3. #3
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    Re: Probability word problem

    Where did you get this problem? It is strange that you should be given such a problem without having learned anything about problems like this. As romsek said, this is a pretty standard "binomial distribution" problem. Do you know what that is? Even if you do not, it is not too difficult to 'reason it out'. The probability that the first 8 people chosen are happy is 0.63^8. The probability the other two are not is 0.37^2. The probability the first 8 are happy and the last two are not is (0.63^8)(0.37^2). In fact, it is not difficult to see that the probability that 8 are happy and 2 are not in any specific order is also (0.63^8)(0.37^2). But there are 10!/8!2!= 45 different orders.

    To find the probability that "at least 2" are happy, find the probability that none are happy and that 1 is happy, add to find the probability that either none or 1 is happy then subtract from 1.
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    Re: Probability word problem

    Quote Originally Posted by HallsofIvy View Post
    (0.63^8)(0.37^2)
    Are those exponents?

    (0.63^8) = 0.02481557802?

    (0.37^2) = 0.1369?
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  5. #5
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    Re: Probability word problem

    Quote Originally Posted by romsek View Post
    this is a binomial distribution with parameters $p=0.63,~n=10$

    can you solve it now?
    0.1528764?
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  6. #6
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    Re: Probability word problem

    Quote Originally Posted by MathLearner2016 View Post
    Are those exponents?

    (0.63^8) = 0.02481557802?

    (0.37^2) = 0.1369?
    Yes, those are correct. Now multiply them together to get the probability of any specific ordering of 8 and 2 things, then multiply by 45 to get the probability of all orders.
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  7. #7
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    Re: Probability word problem

    Quote Originally Posted by MathLearner2016 View Post
    0.1528764?
    Yes, to 7 decimal places 45(.63)^{8}(.37)^2= 0.1528764. Why that specific accuracy? Since you are given the probability to only two decimal places, I would have written 0.15.
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  8. #8
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    Re: Probability word problem

    b)

    So the probability of nobody being happy is 0.00005; one being happy is 0.0008 and two being happy is 0.0063. Can I use that to figure out the probability of at least 2 being happy?
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