2017 AMC 12B Problems/Problem 22

Problem 22

Abby, Bernardo, Carl, and Debra play a game in which each of them starts with four coins. The game consists of four rounds. In each round, four balls are placed in an urn---one green, one red, and two white. The players each draw a ball at random without replacement. Whoever gets the green ball gives one coin to whoever gets the red ball. What is the probability that, at the end of the fourth round, each of the players has four coins?

$\textbf{(A)}\quad \dfrac{7}{576} \qquad \qquad \textbf{(B)}\quad \dfrac{5}{192} \qquad\qquad \textbf{(C)}\quad \dfrac{1}{36} \qquad\qquad \textbf{(D)}\quad \dfrac{5}{144} \qquad\qquad\textbf{(E)}\quad \dfrac{7}{48}$

Solution

It amountsI choices for the -1in RA, 2 choices for the rest of 1 -1 in RB, 2 choices between C and D to fill in the remaining pair of 1 and -1. Then double for symmetrical case of <RA>=3 and =1. Total 24 subcases.

Case 3. <RA>=1, and =1. 3 Choices for -1in RA. Then (1)1 in RB goes directly under 1 in RA. Then 2 choices to assign remaining two pairs of 1, -1in RC and RD. Or (2) 1 in RB goes not directly under 1 in RA: 2 choices here. Then 2 choices to finish. Hence 3(2+2×2)=18.

In sum, 3+24+18=45 cases. Probability= 45÷(12^3)=5/192.

2017 AMC 12B (Problems • Answer Key • Resources)
Preceded by Problem 21	Followed by Problem 23
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2017 AMC 12B Problems/Problem 22

Problem 22

Solution

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