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------------------------------------------------ An ice chest contains 6 cans of apple juice, 7 cans of grape juice, 5 cans of orange juice, and 3 cans of pineapple juice. Suppose that you reach into the container and randomly select three cans in succession. What is the probability of selecting no grape juice?

A. 27/30

B. 33/104

C. 11/30

D. 23/60

The student is asking about the probability of randomly selecting three cans from an ice chest and getting no grape juice cans in the process. To find this probability, we need to use combinatorics principles to determine the number of favorable outcomes divided by the total number of possible outcomes. There are 6 cans of apple juice, 7 cans of grape juice, 5 cans of orange juice, and 3 cans of pineapple juice, totaling 21 cans.

To find the number of ways to select three non-grape juice cans, we compute the combination of the remaining cans (6 apple + 5 orange + 3 pineapple = 14 non-grape cans) taken 3 at a time: C(14, 3). The total number of ways to select any three cans from the 21 is C(21, 3).

The probability of selecting no grape juice is then:

P(no grape juice) = C(14, 3) / C(21, 3)

P(no grape juice) = (14! / (3! * (14 - 3)!)) / (21! / (3! * (21 - 3)!))

After simplification, we get:

P(no grape juice) = (14 * 13 * 12) / (21 * 20 * 19)

This reduces to:

P(no grape juice) = 11/30

Therefore, the correct answer is C. 11/30.