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Probability of choosing stale donuts out of 24

  1. Sep 2, 2016 #1
    1. The problem statement, all variables and given/known data
    There are 6 stale donuts in a set of 24. What is the probability of:
    a) there being no stale donuts in a sample of 10?
    b) there being 3 stale donuts in a sample of 10?
    c) What is the chance of a stale doughnut being found?

    2. Relevant equations
    [itex]N \, permutations = N![/itex]

    3. The attempt at a solution
    Let ##X## denote the number of stale donuts in a set of 10.

    a) I used the idea of permutations like so:
    [tex]P(X = 0) = \frac{\frac{18!}{8!}}{\frac{24!}{15!}} \approx 0.335[/tex]
    This was incorrect, according to the testing software

    b) Here I followed the same idea:
    [tex]P(X=3) = \frac{\frac{18!}{11!} \times \frac{6!}{3!}}{\frac{24!}{15!}} \approx 0.041[/tex]

    c) This is just the complement of part ##a##:

    [tex]P(X \, is \, at \, least \, 1) = 1 - 0.061 = 0.665[/tex]

    Parts ##a## and ##b## (and ##c## as a consequence of ##a## being wrong) are apparently wrong and I'm not sure whats wrong with my reasoning.
     
    Last edited: Sep 2, 2016
  2. jcsd
  3. Sep 2, 2016 #2
    I was completely wrong. I was supposed to use combinations instead:

    a) [tex]P(X = 0) = \frac{18 \choose 10}{24 \choose 10} =39/1748 \approx 0.22 [/tex]

    b) [tex]P(X = 3) = \frac{{18 \choose 7} \times {6 \choose 3}}{{24 \choose 10}} =1560/4807 \approx 0.325 [/tex]

    c) [tex]P(X \text{ is at least } 1) = 1 - P(X = 0) = 1709/1748 \approx 0.987[/tex]
     
  4. Sep 2, 2016 #3
  5. Sep 2, 2016 #4

    Ray Vickson

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    Homework Helper

    These are all correct, now. As Buzz Bloom has stated, you are using the so-called hypergeometric distribution.
     
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