Question

If $X$ follows a binomial distribution with parameters $n=8$ and $p=\frac{1}{2}$, then $P(|X-4|\le 2)=$

• A. $\frac{119}{128}$
• B. $\frac{116}{128}$
• C. $\frac{29}{128}$
• D. None of these

Answer 1

The correct option is A $\frac{119}{128}$

We have $P(|X-4|\le 2)=P(-2\le X-4\le 2)=P(2\le X\le 6)$
$=1-[P(X=0)+P(X=1)+P(X=7)+P(X=8)]$

$=1-\left[{}^8{C}_0{\left(\frac{1}{2}\right)}^8{+}^8{C}_1{\left(\frac{1}{2}\right)}^8{+}^8{C}_7{\left(\frac{1}{2}\right)}^8{+}^8{C}_8{\left(\frac{1}{2}\right)}^8\right]$
$=1-{\left(\frac{1}{2}\right)}^8(1+8+8+1)=1-\frac{18}{2^8}=\frac{119}{128}$