Question

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

• A. $\frac{121}{128}$
• B. $\frac{119}{128}$
• C. $\frac{117}{128}$
• D. $\frac{115}{128}$

Answer 1

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

Here,$P=\frac{1}{2},n=8$
$\therefore q=1-p=1-\frac{1}{2}=\frac{1}{2}$
$\therefore$ The binomial distribution is ${(\frac{1}{2}+\frac{1}{2})}^8$
Also, $|x-4|\le 2$
$\Rightarrow -2\le x-4\le 2\Rightarrow 2\le x\le 6\therefore p(|x-4|\le 2)=p(x=2)+p(x=3)+p(x=4)+p(x-5)+p(x=6)$
${=}^8{C}_2{\left(\frac{1}{2}\right)}^6{\left(\frac{1}{2}\right)}^2{+}^8{C}_3{\left(\frac{1}{2}\right)}^5{\left(\frac{1}{2}\right)}^3{+}^8{C}_4{\left(\frac{1}{2}\right)}^4{\left(\frac{1}{2}\right)}^4{+}^8{C}_6{\left(\frac{1}{2}\right)}^3{\left(\frac{1}{2}\right)}^5{+}^8{C}_6{\left(\frac{1}{2}\right)}^2{\left(\frac{1}{2}\right)}^6$
$=\frac{{}^8{C}_2{+}^8{C}_3{+}^8{C}_4{+}^8{C}_5{+}^8{C}_6}{2^8}$
=$\frac{238}{256}=\frac{119}{128}$