AMC10 2006 A

AMC10 2006 A · Q13

AMC10 2006 A · Q13. It mainly tests Probability (basic), Conditional probability (basic).

A player pays $5 to play a game. A die is rolled. If the number on the die is odd, the game is lost. If the number on the die is even, the die is rolled again. In this case the player wins if the second number matches the first and loses otherwise. How much should the player win if the game is fair? (In a fair game the probability of winning times the amount won is what the player should pay.)

玩家支付 5 美元玩一个游戏。掷一个骰子。如果骰子上的数字是奇数，游戏输掉。如果是偶数，再掷一次骰子。在这种情况下，如果第二次数字与第一次相同则赢，否则输。如果游戏公平，玩家应该赢得多少？（在公平游戏中，赢的概率乘以赢得金额等于玩家支付的金额。）

(A) $12 $12

(B) $30 $30

(D) $60 $60

(E) $100 $100

Answer

Correct choice: (D)

正确答案：（D）

Solution

(D) Let $x$ represent the amount the player wins if the game is fair. The chance of an even number is $1/2$, and the chance of matching this number on the second roll is $1/6$. So the probability of winning is $(1/2)(1/6)=1/12$. Therefore \((1/12)x=\$5 and x=\$60.

(D) 设 $x$ 表示如果游戏公平，玩家赢得的金额。掷出偶数的概率是 $1/2$，第二次掷出与该数相同的概率是 $1/6$。因此获胜的概率是 $(1/2)(1/6)=1/12$。所以 $(1/12)x=\$5，且 \(x=\$60$。

Topics