Let $\mathrm{X} \sim \mathrm{B}\left(6, \frac{1}{2}\right)$, then $\mathrm{P}[|x-4| \leqslant 2]$ is

Let $\mathrm{X} \sim \mathrm{B}\left(6, \frac{1}{2}\right)$, then $\mathrm{P}[|x-4| \leqslant 2]$ is
  1. $\frac{115}{128}$
  2. $\frac{63}{64}$
  3. $\frac{57}{64}$
  4. $\frac{7}{64}$

Solution

$\mathrm{X} \sim \mathrm{~B}\left(6, \frac{1}{2}\right)$
Here, $n=6, p=\frac{1}{2}, q=\frac{1}{2}$ Consider, $|x-4| \leq 2$ $\begin{array}{ll} \therefore & -2 \leq x-4 \leq 2 \\ \therefore & 2 \leq x \leq 6 \\ \therefore \quad & \mathrm{P}[|x-4| \leq 2]=\mathrm{P}(2 \leq x \leq 6) \\ & =1-[\mathrm{P}(\mathrm{X}=0)+\mathrm{P}(\mathrm{X}=1)] \\ & =1-\left[{ }^6 \mathrm{C}_0 \mathrm{p}^0 \mathrm{q}^6+{ }^6 \mathrm{C}_1 \mathrm{p}^1 \mathrm{q}^5\right] \\ & =1-\left[\frac{1}{2^6}+\frac{6}{2^6}\right] \\ & =1-\frac{7}{64}=\frac{57}{64} \end{array}$

Asked in: MHT CET 2024 (03 May Shift 1)

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