In Ms. Q's deck of cards, every card is one of four colors (red, green, blue, and yellow), and is labeled with one of seven numbers (1, 2, 3, 4, 5, 6, and 7). Among all the cards of each color, there is exactly one card labeled with each number. The cards in Ms. Q's deck are shown below.
Yunseol draws 5 cards from Ms. Q's deck. What is the probability that three cards have the same number?
What is the probability that three cards have the same number?
Descriptive solution below:
+2487
Solution:
Remove slop solution ... replace with a correct solution.
Case analysis of binomial distribution.
\(\huge \rho(x) \normalsize = \dfrac { \left [\dbinom {7}{1} \dbinom {4}{3} \right] * \left[ \left (\dbinom {6}{2} \dbinom {4}{1}\dbinom {4}{1} \right ) + \left(\dbinom {6}{1} \dbinom {4}{2}\right) \right]} {\dbinom {28}{5}} = \underbrace {\left (\large \dfrac {46}{585}\right)}_{\text {exact probability}} \left ( \approx 7.86\% \right )\\ \)
Expansion, Dissection, and Description:
nCr(7,1)*nCr(4, 3) * nCr(6,2)*nCr(4, 1)*nCr(4, 1) = 6720 |Counts of triples with non-pairs.
Choose one (1) of seven (7) numbers; choose three (3) of the four (4) colors; choose two (2) of the six (6) remaining numbers; choose one (1) of four colors for each of the two (2) numbers.
nCr(7,1)*nCr(4, 3) * (nCr(6,1)*nCr(4, 2) = 1008 |Counts of triples and only pairs.
Choose one (1) of seven (7) numbers; choose three (3) of the four (4) colors; choose one (1) of the six (6) remaining numbers; choose two (2) of the four (4) colors –making a pair of numbers.
Add these counts: 6720 + 1008 = 7728
Divide: 7728/ nCr(28,5) = 46/585 = 7.86%
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See also Tiggsy's solutions
GA
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