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| − | | + | #redirect[[2023 AMC 12B Problems/Problem 19]] |
| − | == 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>
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