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Probability of picking someone who opposes and favors

  1. Apr 11, 2012 #1
    Out of 25 people, 17 favor and 8 oppose something. What is the probability of picking two who favor and four who oppose out of six people?

    25C6=177100 to find the # of ways to get 6 people.
    Then find those who favor 17C2=136
    Then find those who oppose 8C4=70
    Multiply that together 136×70=9520
    Then divide by total number of ways to pick 6 people 9520/177100=.0537549407
    Did I do this right?
     
  2. jcsd
  3. Apr 11, 2012 #2

    Ray Vickson

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    Yes, you did just fine. If you want to look at it more generally, this problem involves a standard hypergeometric distribution, defined as follows. If we have a population of N1 things of type 1 and N2 things of type 2, and pick a sample of n things at random without replacement, the probability of getting k type 1 things in the sample is
    [tex] P\{ k \text{ type 1 }\} = \frac{C(N_1,k) C(N_2,n-k)}{C(N_1 + N_2,n)},[/tex]
    where C(a,b) is the binomial coefficient
    [tex] C(a,b) = \frac{a!}{b! (a-b)!} = \frac{1(a-1) \cdots (a-b+1)}{b!}. [/tex]
    In your case you have N1 = 17, N2 = 8, n = 6 and k = 2, so the probability is
    [tex] p = \frac{C(17,2)C(8,4)}{C(25,6)} = \frac{17\cdot 16}{2 \cdot 1} \cdot \frac{8 \cdot 7 \cdot 6 \cdot 5}{4 \cdot 3 \cdot 2 \cdot 1} \cdot
    \frac{6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1}{25 \cdot 24 \cdot 23 \cdot 22 \cdot 21} = \frac{68}{1265} \doteq 0.05375 . [/tex]

    RGV
     
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