An urn contains 5 red and 6 blue and 8 green balls. 3 balls are randomly selected from the urn, find the probability that they are all of the different colors if the balls are drawn without replacement
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$\begingroup$ What are your thoughts on this problem? $\endgroup$– paw88789Feb 10, 2015 at 20:27
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$\begingroup$ 'they are all of the different colors' - it's not entirely clear if outcomes RGB and RBG should be counted as one or not $\endgroup$– AlexFeb 10, 2015 at 20:34
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3 Answers
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Since $19$ is the total number of balls, then there are ${19 \choose 3}$ ways to choose 3 balls if you ignore the ordering of the balls. The number of ways to get 3 balls of different color (ignoring the ordering) is $5 \cdot 6 \cdot 8$. Does that help?
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$${\rm P}=\frac{^5{\rm C}_1^6{\rm C}_1^8{\rm C}_1}{^{19}{\rm C}_3}$$
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Using the given answers the probability is $\frac{8000}{323}\%\approx 24.8\%$
