1. $\frac{1}{4}$
  2. $\frac{1}{8}$
  3. $\frac{3}{2}$
  4. $\frac{1}{2}$

Solution

Here, $\mathrm{K}+\frac{1}{6}+\frac{3}{8}+2 \mathrm{~K}+\frac{1}{12}=1$ $\begin{aligned} & \Rightarrow 3 K+\frac{4+9+2}{24}=1 \\ & \Rightarrow 3 K=1-\frac{15}{24}=\frac{9}{24}=\frac{1}{8}\end{aligned}$

Asked in: MHT CET 2022 (05 Aug Shift 2)

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