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Difference between revisions of "2023 AMC 10B Problems/Problem 21"

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== Solution ==
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We first examine the possible arrangements for parity of number of balls in each box for <math>2022</math> balls.
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If a <math>0</math> denotes an even number and a <math>1</math> denotes an odd number, then the distribution of balls for <math>2022</math> balls could be <math>000,011,101,</math> or <math>110</math>. With the insanely overpowered magic of cheese, we assume that each case is about equally likely.
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From <math>000</math>, it is not possible to get to all odd by adding one ball; we could either get <math>100,010,</math> or <math>001</math>. For the other <math>3</math> cases, though, if we add a ball to the exact right place, then it'll work.
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For each of the working cases, we have <math>1</math> possible slot the ball can go into (for <math>101</math>, for example, the new ball must go in the center slot to make <math>111</math>) out of the <math>3</math> slots, so there's a <math>\dfrac13</math> chance. We have a <math>\dfrac34</math> chance of getting one of these working cases, so our answer is <math>\dfrac34\cdot\dfrac13=\boxed{\textbf{(E) }\dfrac14.}</math>

Revision as of 14:42, 15 November 2023

Solution

We first examine the possible arrangements for parity of number of balls in each box for $2022$ balls.

If a $0$ denotes an even number and a $1$ denotes an odd number, then the distribution of balls for $2022$ balls could be $000,011,101,$ or $110$. With the insanely overpowered magic of cheese, we assume that each case is about equally likely.

From $000$, it is not possible to get to all odd by adding one ball; we could either get $100,010,$ or $001$. For the other $3$ cases, though, if we add a ball to the exact right place, then it'll work.

For each of the working cases, we have $1$ possible slot the ball can go into (for $101$, for example, the new ball must go in the center slot to make $111$) out of the $3$ slots, so there's a $\dfrac13$ chance. We have a $\dfrac34$ chance of getting one of these working cases, so our answer is $\dfrac34\cdot\dfrac13=\boxed{\textbf{(E) }\dfrac14.}$

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