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A deck of cards consists of 8 blue cards and 5 white cards. A simple random sample (random draws without replacement) of 6 cards is selected.

What is the chance that one of the colors appears twice as many times as the other?

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    if you do: x+2x=6 you get that x=2 and 2x=4 is the only possible combination of colors that fulfills "one of the colors appears twice as many times as the other",then you need to analyze:
    2 BLUE & 4 WHITE (case 1) or 2 WHITE & 4 BLUE (case 2)
    We need to apply hypergeometric distribution here:
    p(k)=(K C k)(N−K C n−k)/(N C n)
    Where (I will put the numbers for case 1):
    K=Maximum number of successes (number of BLUE cards=8)
    k=number of expected successes (2 BLUE)
    N=Population size (13 cards=8 BLUE+5 WHITE)
    n=number of draws(6).
    Then you have for the first case:.
    p1(k=2)=(8 C 2)(5 C 4)/ (13 C 6)=0.0815

    case "2 WHITE+4 BLUE" P(k2=2) = 0.4079

    It is case 1 or case 2, then the total probability is 0.0815+0.4079=0.4895, that is a 48.95%

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