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Four red balls, four green balls, and four blue balls are put in a box. Three balls are drawn one after another at random without replacement. The probability that all three balls are red is
- 1/72
- 1/55
- 1/36
- 1/27
Correct answer: 1/55
Solution
The probability of drawing three red balls in succession without replacement is \(\frac{4}{12}\cdot\frac{3}{11}\cdot\frac{2}{10}\). This equals \(\frac{24}{1320}=\frac{1}{55}\).
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