Question

If $x\in [0,8]$, the probability that $x^2-8x+12\ge 0$ is

• A. $\frac{1}{2}$
• B. $3/4$
• C. $\frac{1}{4}$
• D. None of these

Answer 1

Answer: (A)

Length of interval $=8-0=8$
Also $(x-2)(x-6)\ge 6$
Using Wavy curve method, we get
$x\le 2$ or $x\ge 6$
$\Rightarrow$ Interval length for favourable points ,br> $=2-0+(8-6)=4$
$\therefore$ Required probability $=\frac{4}{8}=1/2$
Short Cut Method :
$x^2-8x+12\ge 0$
$\Rightarrow (x-2)(x-6)\ge 0$
$\Rightarrow x\le 2$ or $x\ge 6$
$\therefore$ Required probability $=\frac{\underset{0}{\overset{2}{\int }}dx+\underset{6}{\overset{8}{\int }}dx}{\underset{0}{\overset{8}{\int }}dx}$
$=\frac{2+2}{8}=1/2$