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费城动物园

Philadelphia Zoo

专题
Probability / 概率
难度
L3

题目详情

在费城动物园的 Water is Life 展区，有 2 种动物：巨獭和小熊猫。每只动物都会做以下 5 种动作之一：进食、睡觉、打滚、游泳、休息。我在展区里看到了两只动物，其中一只是正在游泳的小熊猫。假设我看到物种与动作的任意组合（可重复）都等可能，那么两只动物都在游泳的概率是多少？

At the Philadelphia Zoo, there are 2 types of animals at the Water is Life exhibit: giant otter, red panda. The animals each perform one of these 5 actions: eating, sleeping, rolling, swimming, resting. I saw two animals at the exhibit and one of them was a red panda swimming. Assuming that it is equally likely for me to have seen any combination of species and actions (including duplicates), what is the probability that both animals were swimming?

解析

我们按以下步骤求解：

步骤 1：计算有序结果总数

每只动物有 2 种物种 × 5 种动作 = 10 种可能的 (species, action) 状态。对两只动物（有序）而言，总结果数为： ∣Ω∣=10×10=100.|Ω| = 10 × 10 = 100.

步骤 2：事件 B："至少一只动物是 (R, swimming)"

令 RS 表示特定组合（小熊猫，游泳）。

两只动物都不是 RS 的有序对数量为 9 × 9 = 81（每只动物都必须是其余 9 种组合之一）。

因此，至少有一只是 RS 的有序对数量为： ∣B∣=100−81=19.|B| = 100 - 81 = 19.

步骤 3：事件 S："两只动物都在游泳（action = swimming）"

若一只动物在游泳，它的物种可以是 G 或 R。因此每只动物都有 2 种可能：(G,swim) 或 (R,swim)。对于两只有序动物： ∣S∣=2×2=4.|S| = 2 × 2 = 4. 这 4 个有序对是： (G,swim; G,swim), (G,swim; R,swim), (R,swim; G,swim), (R,swim; R,swim).

步骤 4：交集 S ∩ B：两只都在游泳且至少一只是 RS。

在这 4 个“双双游泳”的有序对中，唯一不包含 RS 的是 (G,swim; G,swim)。因此： ∣S∩B∣=4−1=3.|S ∩ B| = 4 - 1 = 3.

步骤 5：条件概率

我们要求： P(both swimming | at least one is RS)=∣S∩B∣/∣B∣=3/19.数值结果： 3/19\text{P(both swimming | at least one is RS)} = |S ∩ B| / |B| = 3 / 19. \\ \text{数值结果：} \ \boxed{3/19}

Original Explanation

We'll walk through the solution using the following steps:

Step #1: Count total ordered outcomes

Each animal has 2 species × 5 actions = 10 possible (species,action) states. For two animals (ordered), total outcomes: ∣Ω∣=10×10=100.|Ω| = 10 × 10 = 100.

Step #2: Event B: “at least one animal is (R, swimming)”

Let RS denote the specific combination (red panda, swimming).

Number of ordered pairs with no RS on either animal: 9 × 9 = 81 (each animal must be one of the 9 other combos).

Thus number of ordered pairs with at least one RS: ∣B∣=100−81=19.|B| = 100 − 81 = 19.

Step #3: Event S: “both animals are swimming (action = swimming)”

For each animal to be swimming, its species may be either G or R. So each animal has 2 possibilities (G,swim) or (R,swim). For two ordered animals: ∣S∣=2×2=4.|S| = 2 × 2 = 4. These 4 ordered pairs are: (G,swim; G,swim), (G,swim; R,swim), (R,swim; G,swim), (R,swim; R,swim).

Step #4: Intersection S ∩ B: both swimming and at least one is RS.

From the 4 both-swimming ordered pairs, the only one that does not contain RS is (G,swim; G,swim). So: ∣S∩B∣=4−1=3.|S ∩ B| = 4 − 1 = 3.

Step #5: Conditional probability

We want: P(both swimming | at least one is RS)=∣S∩B∣/∣B∣=3/19.Numeric value: 3/19\text{P(both swimming | at least one is RS)} = |S ∩ B| / |B| = 3 / 19. \\ \text{Numeric value:} \ \boxed{3/19}