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2018 AMC 12A Problems/Problem 24

Revision as of 17:58, 8 February 2018 by Random guy (talk | contribs) (Undo revision 90663 by Random guy (talk))

Problem

Alice, Bob, and Carol play a game in which each of them chooses a real number between 0 and 1. The winner of the game is the one whose number is between the numbers chosen by the other two players. Alice announces that she will choose her number uniformly at random from all the numbers between 0 and 1, and Bob announces that he will choose his number uniformly at random from all the numbers between $\tfrac{1}{2}$ and $\tfrac{2}{3}.$ Armed with this information, what number should Carol choose to maximize her chance of winning?


$\textbf{(A) }\frac{1}{2}\qquad \textbf{(B) }\frac{13}{24} \qquad \textbf{(C) }\frac{7}{12} \qquad \textbf{(D) }\frac{5}{8} \qquad \textbf{(E) }\frac{2}{3}\qquad$

Solution 1

Plug in all the answer choices to get $\boxed{\textbf{(B)}.}$

Solution 2

Let the value we want be $x$. The probability that Alice's number is less than Carol's number and Bob's number is greater than Carol's number is $x(\frac{2}{3}-x)$. Similarly, the probability that Bob's number is less than Carol's number and Alice's number is greater than Carol's number is $(x-\frac{1}{2})(1-x)$. Adding these together, the probability that Carol wins given a certain number $x$ is $-2x^2+\frac{13}{6}x-\frac{1}{2}$. Using calculus or the fact that the extremum of a parabola occurs at $\frac{-b}{2a}$, the maximum value occurs at $x=\frac{13}{24}$, which is $\boxed{\textbf{(B)}.}$

Solution 3

The expected value of Alice's number is $\frac{1}{2}$ and the expected value of Bob's number is $\frac{7}{12}$. To maximize her chance of winning, Carol would choose number exactly in between the two expected values, giving:$\frac{6+7}{12*2}=\frac{13}{24}$. This is $\boxed{\textbf{(B)}}$. (Random_Guy)

See Also

2018 AMC 12A (Problems • Answer Key • Resources)
Preceded by
Problem 23 		Followed by
Problem 25
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 12 Problems and Solutions

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