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Question: Answered & Verified by Expert
If $X \sim B\left(8, \frac{1}{2}\right)$, then $P(|x-4| \leq 2)=$
MathematicsProbabilityMHT CETMHT CET 2020 (20 Oct Shift 2)
Options:
  • A $\frac{119}{128}$
  • B $\frac{29}{128}$
  • C $\frac{238}{728}$
  • D $\frac{119}{228}$
Solution:
1243 Upvotes Verified Answer
The correct answer is: $\frac{119}{128}$
We have $\mathrm{n}=8, \mathrm{p}=\frac{1}{2} \quad \Rightarrow \mathrm{q}=\frac{1}{2}$
To find $\mathrm{P}(|\mathrm{X}-4| \leq 2) \quad \Rightarrow \mathrm{X}=2,3,4,5,6$
Hence required probability
$$
\begin{array}{l}
={ }^{8} C_{2}\left(\frac{1}{2}\right)^{6}\left(\frac{1}{2}\right)^{2}+{ }^{8} C_{3}\left(\frac{1}{2}\right)^{5}\left(\frac{1}{2}\right)^{3}+{ }^{8} C_{4}\left(\frac{1}{2}\right)^{4}\left(\frac{1}{2}\right)^{4}+{ }^{8} C_{5}\left(\frac{1}{2}\right)^{3}\left(\frac{1}{2}\right)^{5}+{ }^{8} C_{6}\left(\frac{1}{2}\right)^{2}\left(\frac{1}{2}\right)^{6} \\
=\left(\frac{1}{2}\right)^{8}\left[{ }^{8} C_{2}+{ }^{8} C_{3}+{ }^{8} C_{4}+{ }^{8} C_{5}+{ }^{8} C_{6}\right] \\
=\frac{1}{256}[28+56+70+56+28]=\frac{238}{256}=\frac{119}{128}
\end{array}
$$

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