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Choosing a distribution

  1. Apr 14, 2009 #1
    I have been given a question that reads:
    In a taste testing experiment N judges are offered four different brands of orange juice to taste.
    Suppose Y is the number of of judges that choose brand A as the best.

    I know that the probability of a judge choosing A is 1/4.

    I have been asked what is the probability that at least one judge chooses brand A given that there are 10 judges.

    My question is what would be the best distribution to use?

    As I'm going to have to simulate 1000 trials.

    Thanks
    Brendan
     
  2. jcsd
  3. Apr 14, 2009 #2
    No need to simulate. Think of it this way... what is the probability that NO judge chooses A?
     
  4. Apr 14, 2009 #3
    The probability of not A is 0.75 but that is for 1 judge, but the probability of 10 judges not picking juice A would be a lot lower.

    I thought I might be able to use a binomial distribution to figure it out so I can simulate for any number of judges?
    regards
    Brendan
     
  5. Apr 14, 2009 #4
    Just wondering if this makes sense.

    Let x = 1 be the number of successes (here the success is that the judges don't pick A)
    Let n = the number of judges = 10

    So P (x) = 0.75

    If I use the binomial distribution

    P(x=0) = 10Choose0 * P(x)^x * P(not x)^n-x

    P(x=0) = 10Choose1 * (3/4)^0 * (1/4)^10

    P(x=0) = 0.000000953

    In another words it would be extremely improbable to have no judge pick A
     
  6. Apr 14, 2009 #5
    So would the probability of at least one picking juice A be P(1-P(x=0) = 0.000000953)

    > 99.9

    regards
    Brendan
     
  7. Apr 14, 2009 #6
    Sorry guy's I think I've got it wrong.
    I'll try again!

    Let x = 1 be the number of successes (here the success is that the judges pick A)
    Let n = the number of judges = 10

    So P (x) = 0.25

    If I use the binomial distribution to find the probabilty of 10 judges not choosing A

    P(x=0) = 10Choose0 * P(x)^x * P(not x)^n-x

    P(x=0) = 10Chooseo * (1/4)^0 * (3/4)^10

    P(x=0) = 0.056

    So the probabilty of at least 1 must be (1 - 0.056) = 0.943

    I hope this is right cause I'm loosing my mind!!!
    Brendan
     
  8. Apr 15, 2009 #7
    Yep, 0.943 is right.
     
  9. Apr 15, 2009 #8
    Thanks mate!
     
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