A biased die is marked with numbers 2 , 4 , 8 , 16 , 32 , 32 on its faces and the probability of getting a…

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

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

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

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