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Consider a cafeteria with three meals. For a randomly selected patron, it is known that 30% of patrons get meal A,30% of patrons get meal B,40% of patrons get meal C.Suppose we randomly select two patrons. What is the probability that these people have the same meal? Calculate your answer to 4 decimal places.

Question

Consider a cafeteria with three meals. For a randomly selected patron, it is known that 30% of patrons get meal A,30% of patrons get meal B,40% of patrons get meal C.Suppose we randomly select two patrons. What is the probability that these people have the same meal? Calculate your answer to 4 decimal places.

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Solution

To solve this problem, we need to consider the probability of each meal being chosen by both patrons.

  1. The probability that both patrons choose meal A is (0.30) * (0.30) = 0.09 or 9%.
  2. The probability that both patrons choose meal B is (0.30) * (0.30) = 0.09 or 9%.
  3. The probability that both patrons choose meal C is (0.40) * (0.40) = 0.16 or 16%.

To find the total probability that both patrons choose the same meal, we add up these probabilities:

0.09 (meal A) + 0.09 (meal B) + 0.16 (meal C) = 0.34 or 34%.

So, the probability that two randomly selected patrons have the same meal is 0.34 or 34%. To four decimal places, this is 0.3400.

This problem has been solved

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