2024 AMC 12B Problem 25

Each ball is independently one of four equally likely types: red-striped, red-dotted, blue-striped, blue-dotted. Suppose among the $6$ balls there are $k$ red-striped, with $R$ red and $S$ striped in total. For Frida’s uniform pick, $P(\text{red}) = \tfrac{R}{6},$ $P(\text{striped}) = \tfrac{S}{6},$ and $P(\text{red and striped}) = \tfrac{k}{6}.$ Independence means $\dfrac{k}{6} = \dfrac{R}{6}\cdot\dfrac{S}{6},$ i.e. $6k = RS.$ First count assignments with $R\in\{0,6\}$ or $S\in\{0,6\}.$ Each of these four conditions leaves $2^6=64$ choices for the other attribute, and the four assignments at their pairwise intersections have been counted twice. Their union therefore contributes $4\cdot64-4=252.$ For $1\le R,S\le5,$ the condition that $RS$ be divisible by $6$ leaves only $(R,S)=(2,3),(3,2)$ or $(3,4),(4,3).$ In each case $k=\frac{RS}{6},$ and the four type counts are a permutation of $(1,1,2,2).$ Thus each pair contributes $\dfrac{6!}{1!1!2!2!}=180$ assignments. The favorable count is therefore $252+4\cdot180=972.$ Out of $4^6=4096$ equally likely assignments, the probability is $\dfrac{972}{4096}=\dfrac{243}{1024},$ so $m=243.$ Thus, the correct answer is A.