Problem

A refrigerator contains 3 cans of diet soda and 4 cans of regular soda. If two cans of soda are randomly selected without replacement, determine the probability that both cans selected are diet soda. Use combinations to determine the probability.

The probability that both cans selected are diet soda is
(Type an integer or a simplified fraction.)

Answer

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Answer

Final Answer: The probability that both cans selected are diet soda is \(\boxed{\frac{1}{7} \approx 0.143}\).

Steps

Step 1 :We have a refrigerator that contains 3 cans of diet soda and 4 cans of regular soda. We are to determine the probability that if two cans of soda are randomly selected without replacement, both cans selected are diet soda.

Step 2 :We can use combinations to determine the probability. The total number of sodas, \(n\), is 7 (3 diet sodas + 4 regular sodas). The number of diet sodas, \(k\), is 3. The number of sodas to choose, \(r\), is 2.

Step 3 :We first calculate the total number of ways to draw 2 sodas from 7, which is given by the combination \(C(n, r)\). Using the combination formula, we find that there are \(C(7, 2) = 21\) ways.

Step 4 :Next, we calculate the number of ways to draw 2 diet sodas from 3, which is given by the combination \(C(k, r)\). Using the combination formula, we find that there are \(C(3, 2) = 3\) ways.

Step 5 :The probability of drawing 2 diet sodas is then given by the ratio of the number of ways to draw 2 diet sodas to the total number of ways to draw 2 sodas. This is given by \(\frac{C(k, r)}{C(n, r)} = \frac{3}{21} = \frac{1}{7}\).

Step 6 :Final Answer: The probability that both cans selected are diet soda is \(\boxed{\frac{1}{7} \approx 0.143}\).

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