If x sim Bleft(6, frac{1}{2}right),then p(|x- | vedclass

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(B) Given $x \sim B\left(n=6, p=\frac{1}{2}\right)$.The probability mass function is $P(x=k) = \binom{6}{k} \left(\frac{1}{2}\right)^k \left(\frac{1}{

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$\frac{41}{64}$

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(B) Given $x \sim B\left(n=6, p=\frac{1}{2}\right)$.The probability mass function is $P(x=k) = \binom{6}{k} \left(\frac{1}{2}\right)^k \left(\frac{1}{2}\right)^{6-k} = \binom{6}{k} \left(\frac{1}{2}\right)^6 = \binom{6}{k} \frac{1}{64}$.We need to find $P(|x-2| \leqslant 1)$.$|x-2| \leqslant 1$ implies $-1 \leqslant x-2 \leqslant 1$,which simplifies to $1 \leqslant x \leqslant 3$.So,we need to calculate $P(x=1) + P(x=2) + P(x=3)$.$P(x=1) = \binom{6}{1} \frac{1}{64} = \frac{6}{64}$.$P(x=2) = \binom{6}{2} \frac{1}{64} = \frac{15}{64}$.$P(x=3) = \binom{6}{3} \frac{1}{64} = \frac{20}{64}$.Summing these probabilities: $P(1 \leqslant x \leqslant 3) = \frac{6+15+20}{64} = \frac{41}{64}$.

If $x \sim B\left(6, \frac{1}{2}\right)$,then $p(|x-2| \leqslant 1)=$

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