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A bag contains 6 cherry, 3 orange and 2 lemon candies. You reach into the bag and grab 3 candies at random. What is the probability that you grab 2 cherry and 1 orange candy?

1 Answer

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Final answer:

The probability of grabbing 2 cherry and 1 orange candy from the bag is 3/11.

Step-by-step explanation:

To find the probability of grabbing 2 cherry and 1 orange candy, we need to calculate the favorable outcomes and divide it by the total outcomes.

The total number of candies in the bag is 6 (cherry) + 3 (orange) + 2 (lemon) = 11 candies.

The possible favorable outcomes are: choosing 2 cherry candies from the 6 available cherry candies and 1 orange candy from the 3 available orange candies. We can calculate this using the combination formula:

C(6, 2) * C(3, 1) = (6! / (2!(6-2)!) * (3! / (1!(3-1)!)) = 15 * 3 = 45

The total possible outcomes are choosing 3 candies from the 11 available candies:

C(11, 3) = 11! / (3!(11-3)!) = 165

Therefore, the probability of grabbing 2 cherry and 1 orange candy is 45/165, which simplifies to 3/11.

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