AMC10 2007 B

AMC10 2007 B · Q22

AMC10 2007 B · Q22. It mainly tests Probability (basic), Expected value (basic).

A player chooses one of the numbers 1 through 4. After the choice has been made, two regular four-sided (tetrahedral) dice are rolled, with the sides of the dice numbered 1 through 4. If the number chosen appears on the bottom of exactly one die after it is rolled, then the player wins \$1. If the number chosen appears on the bottom of both of the dice, then the player wins \$2. If the number chosen does not appear on the bottom of either of the dice, the player loses \$1. What is the expected return to the player, in dollars, for one roll of the dice?

一名玩家选择 1 到 4 中的一个数字。选择后，掷两个正四面体骰子，骰子面编号 1 到 4。如果所选数字恰好出现在一个骰子的底部，玩家赢得 $1。如果出现在两个骰子的底部，赢得 $2。如果都不出现，输 $1$。掷一次骰子的玩家的期望收益是多少美元？

(A) $-\frac{1}{8}$ $-\frac{1}{8}$

(B) $-\frac{1}{16}$ $-\frac{1}{16}$

(D) $\frac{1}{16}$ $\frac{1}{16}$

(E) $\frac{1}{8}$ $\frac{1}{8}$

Answer

Correct choice: (B)

正确答案：（B）

Solution

Answer (B): The probability of the number appearing 0, 1, and 2 times is $P(0)=\frac{3}{4}\cdot\frac{3}{4}=\frac{9}{16},\quad P(1)=2\cdot\frac{1}{4}\cdot\frac{3}{4}=\frac{6}{16},\quad \text{and}\quad P(2)=\frac{1}{4}\cdot\frac{1}{4}=\frac{1}{16},$ respectively. So the expected return, in dollars, to the player is $P(0)\cdot(-1)+P(1)\cdot(1)+P(2)\cdot(2)=\frac{-9+6+2}{16}=-\frac{1}{16}.$

答案（B）：该数字出现 0 次、1 次和 2 次的概率分别为 $P(0)=\frac{3}{4}\cdot\frac{3}{4}=\frac{9}{16},\quad P(1)=2\cdot\frac{1}{4}\cdot\frac{3}{4}=\frac{6}{16},\quad \text{以及}\quad P(2)=\frac{1}{4}\cdot\frac{1}{4}=\frac{1}{16}.$ 因此，玩家的期望收益（以美元计）为 $P(0)\cdot(-1)+P(1)\cdot(1)+P(2)\cdot(2)=\frac{-9+6+2}{16}=-\frac{1}{16}.$

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