A biased die is marked with numbers 2, 4, 8, 16, 32, 32 on its...

$P\left(2\right) = \dfrac{1}{2}, P\left(4\right) = \dfrac{1}{4}, P\left(8\right) = \dfrac{1}{8}, P\left(16\right) = \dfrac{1}{16}, P\left(32\right) = \dfrac{1}{32}$

When we get the faces as $8, 8, 32$ then the probability will be $3\left(\dfrac{1}{8} \times \dfrac{1}{8} \times \dfrac{2}{32}\right) = \dfrac{3}{1024}$

When we get the faces as $16, 16, 16$ then the probability will be $\dfrac{1}{16} \times \dfrac{1}{16} \times \dfrac{1}{16} = \dfrac{1}{4096}$

Hence, the required probability will be$\dfrac{3}{1024} + \dfrac{1}{4096} = \dfrac{13}{4096}$