Answer to Question #184453 in Statistics and Probability for Jane Ramida

Question #184453

There are 15 balls in an urn. 5 red balls, 5 blue balls, and 5 white balls. Three balls are taken simultaneously.




1
Expert's answer
2021-04-27T01:45:55-0400

The order of the balls does not matter so this is a combination problem. For each item, the sample space consists of

C315=15!3!(153)!=13×14×152×3=455C^{15}_3 = \frac{15!}{3!(15-3)!} = \frac{13 \times 14 \times 15}{2 \times 3} = 455 items.


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