Difference between revisions of "2023 AMC 10B Problems/Problem 21"
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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> | 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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Revision as of 13:46, 15 November 2023
Solution
We first examine the possible arrangements for parity of number of balls in each box for 
balls.
If a 
denotes an even number and a 
denotes an odd number, then the distribution of balls for balls could be 
or 
. With the insanely overpowered magic of cheese, we assume that each case is about equally likely.
From 
, it is not possible to get to all odd by adding one ball; we could either get 
or 
. For the other 
cases, though, if we add a ball to the exact right place, then it'll work.
For each of the working cases, we have possible slot the ball can go into (for 
, for example, the new ball must go in the center slot to make 
) out of the slots, so there's a 
chance. We have a 
chance of getting one of these working cases, so our answer is 
~Technodoggo
