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餐厅就餐时段

Restaurant Meal

专题
Probability / 概率
难度
L2

题目详情

三位朋友都决定去一家很受欢迎的餐厅。如果每个人都独立且等概率地随机选择去吃早餐、午餐或晚餐，那么三种就餐时段都被选中的概率是多少？

Three friends all decide to visit a popular restaurant. If each friend randomly and independently chooses to go to either breakfast, lunch, or dinner, what is the probability that all three mealtimes will be selected?

解析

每位朋友都有 3 种选择（早餐、午餐或晚餐）。因为选择彼此独立，所以总的可能结果数为 33=273^3 = 27。

对于有利结果，三位朋友必须分别选择不同的就餐时段。我们来计算这种情况有多少种。

  1. 第一位朋友可以任选 3 个时段中的一个。
  2. 第二位朋友必须选择与第一位不同的时段，因此有 2 种选择。
  3. 第三位朋友必须选择剩下的那个时段，因此有 1 种选择。

所以有利结果数为 3×2×1=63 \times 2 \times 1 = 6。

因此，三位朋友分别选中三个不同就餐时段的概率为： 627=29\frac{6}{27} = \boxed{\frac{2}{9}}

Original Explanation

Each friend has 3 options (breakfast, lunch, or dinner). Since the choices are independent, the total number of possible outcomes (ways the three friends can choose mealtimes) is 33=273^3 = 27.

For the favorable outcomes, each friend must choose a different mealtime. We need to count how many ways this can happen.

  1. The first friend can choose any of the 3 mealtimes.
  2. The second friend must choose a different mealtime from the first, so they have 2 choices.
  3. The third friend must choose the remaining mealtime, so they have 1 choice.

The number of favorable outcomes is 3×2×1=63 \times 2 \times 1 = 6.

The probability that all three friends choose different mealtimes is the ratio of favorable outcomes to total outcomes: 627=29\frac{6}{27} = \boxed{\frac{2}{9}}