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Difference between revisions of "2024 AMC 12B Problems/Problem 25"
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<math>\textbf{(A) } 243 \qquad \textbf{(B) } 245 \qquad \textbf{(C) } 247 \qquad \textbf{(D) } 249\qquad \textbf{(E) } 251</math>
 
<math>\textbf{(A) } 243 \qquad \textbf{(B) } 245 \qquad \textbf{(C) } 247 \qquad \textbf{(D) } 249\qquad \textbf{(E) } 251</math>
  
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==Solution==
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==Solution 1==
  
 
Let <math>a</math> be the number of balls that are both striped and red, <math>x</math> be the number of balls that are striped but blue, and <math>y</math> be the number of balls that are red but dotted. Then there must be <math>6-a-x-y</math> balls that are dotted and blue.
 
Let <math>a</math> be the number of balls that are both striped and red, <math>x</math> be the number of balls that are striped but blue, and <math>y</math> be the number of balls that are red but dotted. Then there must be <math>6-a-x-y</math> balls that are dotted and blue.
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~tsun26
 
~tsun26
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==See also==
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{{AMC12 box|year=2024|ab=B|num-b=24|num-a=-1}}
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{{MAA Notice}}

Revision as of 14:29, 14 November 2024

Problem 25

Pablo will decorate each of $6$ identical white balls with either a striped or a dotted pattern, using either red or blue paint. He will decide on the color and pattern for each ball by flipping a fair coin for each of the $12$ decisions he must make. After the paint dries, he will place the $6$ balls in an urn. Frida will randomly select one ball from the urn and note its color and pattern. The events "the ball Frida selects is red" and "the ball Frida selects is striped" may or may not be independent, depending on the outcome of Pablo's coin flips. The probability that these two events are independent can be written as $\frac mn,$ where $m$ and $n$ are relatively prime positive integers. What is $m?$ (Recall that two events $A$ and $B$ are independent if $P(A \text{ and }B) = P(A) \cdot P(B).$)

$\textbf{(A) } 243 \qquad \textbf{(B) } 245 \qquad \textbf{(C) } 247 \qquad \textbf{(D) } 249\qquad \textbf{(E) } 251$

Solution 1

Let $a$ be the number of balls that are both striped and red, $x$ be the number of balls that are striped but blue, and $y$ be the number of balls that are red but dotted. Then there must be $6-a-x-y$ balls that are dotted and blue.

Let $A$ be the event, "the ball Frida selects is red", and $B$ be the event, "the ball Frida selects is striped". $A$ and $B$ are independent if and only if $\frac{a}{6}=\frac{a+x}{6}\cdot\frac{a+y}{6}$, i.e., \[6a=(a+x)(a+y)\]

Before we continue, I'd like to clarify that the sample space $S$, under which we are computing probability that $A$ and $B$ are independent, consists of $4^6$ total outcomes, each equally likely. Notice that once given $a, x, y$, there are ${6\choose a}{6-a \choose x}{6-a-x \choose y}=\frac{6!}{a!x!y!(6-a-x-y)!}$ corresponding outcomes.

Now, it remains to find all nonnegative integer solutions $(a, x, y)$ to the above equation such that $a+x+y<6$, add up all the corresponding outcomes, and then divide by $4^6$. We can find the solutions relatively easily by doing casework on $a=0, 1, 2, 3, 4, 5, 6$. Long story short, one finds the solutions to be \[(0, 0, 0), (0, 0, 1)\cdots (0, 0, 6)\] \[(0, 1, 0), (0, 2, 0)\cdots (0, 6, 0)\] \[(1, 5, 0), (1, 0, 5), (1, 2, 1), (1, 1, 2)\] \[(2, 4, 0), (2, 0, 4) (2, 2, 1), (2, 1, 2)\] \[(3, 3, 0), (3, 0, 3)\] \[(4, 2, 0), (4, 0, 2)\] \[(5, 1, 0), (5, 0, 1)\] \[(6, 0, 0)\]

There are little tricks we can use to make the process of adding up the corresponding outcomes faster. First, we notice that the first, second, and last row of solutions contributes $2^7=128$ outcomes. Then we can take advantage of the symmetry between $x, y$ to add up the rest of the outcomes. In the end we get \[128+2(6+180+15+180+20+15+6)=972\]

Hence, the probability is $\frac{972}{4^6}=\frac{243}{2^{10}}$. Therefore, the answer is $\fbox{\textbf{(A) } 243}$

~tsun26

See also

2024 AMC 12B (Problems • Answer Key • Resources)
Preceded by
Problem 24 		Followed by
Problem -1
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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