Sunday Afternoon Maths XXXIII

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$$\frac{1}{2}=\mathbb{P}(A=r)\mathbb{P}(B=r)+\mathbb{P}(A=b)\mathbb{P}(B=b)\\ =\frac{5}{6}p+\frac{1}{6}(1-p)\\ =\frac{1}{6}+\frac{4}{6}p $$ $$\frac{4}{6}p=\frac{1}{2}-\frac{1}{6}\\ =\frac{1}{3}\\ p=\frac{\frac{1}{3}}{\frac{4}{6}}=\frac{1}{2}$$

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$$\frac{6729}{134580}$$ $$\frac{7932}{15864}$$ $$\frac{9267}{18534}$$

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