In a sample bag of M&M's candy, there are 5 brown, 6 yellow, 4 blue, 3 green, and 2 orange. What is the probablility of getting 1 brown and 2 orange M&M's if 3 are taken at a time?
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There are 20 candies in all, so the probability of picking a brown one is 5/20=1/4. That leaves 19. The probability of picking an orange one is 2/19, and the probability of picking another orange is 1/18. So if the order were to matter, the combined probability would be 1/4*2/19*1/18. However, the order doesn't matter, and the number of ways of picking a brown and two oranges is: BOO, OBO, OOB, three ways. So we need to multiply the combined probabilities by 3: 3*1/4*2/19*1/18=1/4*1/19*1/3=1/228=0.004386=0.44% approx. and it doesn't make any difference to the probability of picking them one at a time or 3 at once; it just takes less time to pick 3 at once!

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