Question

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

• A. $\frac{117}{128}$
• B. $\frac{118}{128}$
• C. $\frac{119}{128}$
• D. $\frac{63}{64}$

Answer 1

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

$n=8$
$p=\frac{1}{2}$
Since $|X-4|\le 2$
i.e. $-2\le X-4\le 2$
$\Rightarrow 2\le X\le 6$
$P(|X–4|\le 2)=P(2\le X\le 6)$
$\Rightarrow P(X=2)+P(X=3)+P(X=4)+P(X=5)+P(X=6)$
$P(X=r){=}^n{C}_r{p}^r{q}^{n-r}$
$P(2\le X\le 6){=}^8{C}_2{p}^2{q}^6{+}^8{C}_3{p}^3{q}^5{+}^8{C}_4{p}^4{q}^4{+}^8{C}_5{p}^5{q}^3{+}^8{C}_6{p}^6{q}^2$
$=p^8\left[{}^8{C}_2{+}^8{C}_3{+}^8{C}_4{+}^8{C}_5{+}^8{C}_6\right]$Since$p=q$
$=\frac{1}{2^8}[28+56+70+56+28]$
$=\frac{238}{256}=\frac{119}{128}$