20 美元由 5 元与 1 元组成：纸币张数期望

设 $X$ 为 5 美元张数，$Y$ 为 1 美元张数，则

$$5X+Y=20.$$

非负整数解为：

• $X=0, Y=20$（20 张）
• $X=1, Y=15$（16 张）
• $X=2, Y=10$（12 张）
• $X=3, Y=5$（8 张）
• $X=4, Y=0$（4 张）

若每种组合等可能，则

$$\mathbb{E}[X+Y]=\frac{20+16+12+8+4}{5}=12.$$

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Original Explanation

Let:

• $X$ be the number of $5 bills$
• $Y$ be the number of $1 bills$
  Then the total amount of money is: $$5X + Y = 20$$

We are asked to compute the expected value of the total number of bills, i.e.: $$\mathbb{E}[X + Y]$$

Step 1: All valid combinations

We look for all non-negative integer solutions to: $$5X + Y = 20$$

For each value of $X$, $Y = 20 - 5X$ must also be a non-negative integer.

Try all values of $X$ such that $Y \geq 0$:

• $X = 0$, $Y = 20$
• $X = 1$, $Y = 15$
• $X = 2$, $Y = 10$
• $X = 3$, $Y = 5$
• $X = 4$, $Y = 0$

So there are 5 possible combinations:

• $(0, 20)$ → 20 bills
• $(1, 15)$ → 16 bills
• $(2, 10)$ → 12 bills
• $(3, 5)$ → 8 bills
• $(4, 0)$ → 4 bills

Step 2: Expected number of bills

Assume each combination is equally likely (uniform distribution). Then:

$$\mathbb{E}[\text{number of bills}] = \frac{1}{5}(20 + 16 + 12 + 8 + 4) = 12$$