2006 AMC 10A Problems/Problem 13

Problem

A player pays $\textdollar 5$ 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.)

$\textbf{(A) } \textdollar12\qquad\textbf{(B) } \textdollar30\qquad\textbf{(C) } \textdollar50\qquad\textbf{(D) } \textdollar60\qquad\textbf{(E) } \textdollar 100\qquad$

Solution

The probability of rolling an even number on the first turn is $\frac{1}{2}$ and the probability of rolling the same number on the next turn is $\frac{1}{6}$ . The probability of winning is $\frac{1}{2}\cdot \frac{1}{6} =\frac{1}{12}$ . If the game is to be fair, the amount paid, $5$ dollars, must be $\frac{1}{12}$ the amount of the prize money, so the answer is $\boxed{\textbf{(D) } $60}.$

Video Solution

https://youtu.be/IRyWOZQMTV8?t=3410

~ pi_is_3.14

2006 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 12	Followed by Problem 14
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 10 Problems and Solutions

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2006 AMC 10A Problems/Problem 13

Contents

Problem

Solution

Video Solution

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