Q74 / 150 · MHT CET 2024 (03 May Shift 1) · Mathematics
Let $\mathrm{X} \sim \mathrm{B}\left(6, \frac{1}{2}\right)$, then $\mathrm{P}[|x-4| \leqslant 2]$ is
- A$\frac{115}{128}$
- B$\frac{63}{64}$
- Correct answerC$\frac{57}{64}$
- D$\frac{7}{64}$
Solution
$\mathrm{X} \sim \mathrm{~B}\left(6, \frac{1}{2}\right)$
Here, $n=6, p=\frac{1}{2}, q=\frac{1}{2}$ Consider, $|x-4| \leq 2$ $\begin{array}{ll} \therefore & -2 \leq x-4 \leq 2 \\ \therefore & 2 \leq x \leq 6 \\ \therefore \quad & \mathrm{P}[|x-4| \leq 2]=\mathrm{P}(2 \leq x \leq 6) \\ & =1-[\mathrm{P}(\mathrm{X}=0)+\mathrm{P}(\mathrm{X}=1)] \\ & =1-\left[{ }^6 \mathrm{C}_0 \mathrm{p}^0 \mathrm{q}^6+{ }^6 \mathrm{C}_1 \mathrm{p}^1 \mathrm{q}^5\right] \\ & =1-\left[\frac{1}{2^6}+\frac{6}{2^6}\right] \\ & =1-\frac{7}{64}=\frac{57}{64} \end{array}$
Here, $n=6, p=\frac{1}{2}, q=\frac{1}{2}$ Consider, $|x-4| \leq 2$ $\begin{array}{ll} \therefore & -2 \leq x-4 \leq 2 \\ \therefore & 2 \leq x \leq 6 \\ \therefore \quad & \mathrm{P}[|x-4| \leq 2]=\mathrm{P}(2 \leq x \leq 6) \\ & =1-[\mathrm{P}(\mathrm{X}=0)+\mathrm{P}(\mathrm{X}=1)] \\ & =1-\left[{ }^6 \mathrm{C}_0 \mathrm{p}^0 \mathrm{q}^6+{ }^6 \mathrm{C}_1 \mathrm{p}^1 \mathrm{q}^5\right] \\ & =1-\left[\frac{1}{2^6}+\frac{6}{2^6}\right] \\ & =1-\frac{7}{64}=\frac{57}{64} \end{array}$
Hence, the correct answer is option: 'C'.