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Probability two students being chosen

  1. Jul 6, 2013 #1
    1. The problem statement, all variables and given/known data
    From 20 students, 15 will be chosen to enter a tournament. Andy and Tony are among those 20. Find the probability Andy and Tony will be chosen


    2. Relevant equations
    Probability
    Permutation
    Combination


    3. The attempt at a solution
    Probability to choose Andy and Tony = 1 / 20C2

    Probability to choose other 13 = 15C13 / 18C13

    Total probability = 1/20C2 x 15C13 / 18C13

    Am I right?

    Thanks
     
  2. jcsd
  3. Jul 6, 2013 #2

    Ray Vickson

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    This is a "hypergeometric distribution" problem. The original population of ##N = 20## students is split into two parts: (1) a group of ##N_1 = 2## (Andy and Tony); and another group of ##N_2 = 18## (all the others). You want to choose a subset of ##n = 15## students, and you want to know the probability that the subset contains ##k = 2## students from group 1. The general hypergeometric formula (that you can find in textbooks or on-line) is
    [tex] P\{k \text{ type 1} \} = \frac{{N_1 \choose k}{N_2 \choose n-k}}{{N \choose n}} [/tex]
    Does your result look like this for ##N = 20, N_1 = 2, N_2 = 18, n = 15, k = 2##?
     
  4. Jul 6, 2013 #3

    Dick

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    All correct, but you really don't need to look up "hypergeometric", just compute it from first principles. Given that Andy and Tony are in the group, find the number of ways to fill out the group of 15 students. Then divide by the total number of ways to choose 15 students without requiring Andy and Tony be included.
     
    Last edited: Jul 6, 2013
  5. Jul 7, 2013 #4

    Ray Vickson

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    Actually, it is easier than that. First, compute the probability that the group has both type 1s first (that is, either Tony or Andy first) followed by 13 type 2s. This will have the form
    [tex] \frac{2}{20}\frac{1}{19}\frac{18}{18}\frac{17}{17} \cdots \frac{6}{6},[/tex] which is a fraction with ##\text{numerator} = 2 \cdot 1 \cdot 18 \cdot 17 \cdots \cdot 6## and ##\text{denominator} = 20 \cdot 19 \cdot 18 \cdot \cdots \cdot 6##. Any other string with two of type 1 and 13 of type 2 in any other order will have the same numerator and the same denominator (with the same numerator factors all present but in a different order and the denominator factors all present and in the same order), so all strings have the same probability. The whole probability is then the probability of any single string, times the number of distinct strings of the required type; the latter is the binomial coefficient C(15,2) = number of combinations of two things chosen from 15. When you put it all together, you get the hypergeometric formula.
     
  6. Jul 8, 2013 #5
    So, the answer should be: 2C2 . 18C13 / 20C15 ; can you please explain why my logic is wrong?

    Thanks
     
  7. Jul 8, 2013 #6

    Dick

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    I would except I don't understand your logic. You are trying to find the probability Andy and Tony are chosen. The first step in your logic is to state the probability Andy and Tony are chosen is 1/(20C2). I don't follow. You aren't just choosing two people.
     
  8. Jul 9, 2013 #7
    Oh I see my mistake now.

    Thanks for your help, Dick and Ray :smile:
     
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