Let x be a real number randomly chosen so that \(-1 \le x \le 1 \). What is the probability that \(x^2 > \frac12\)?
+118609 If
\(x^2>0.5\\ then\\ x<-\sqrt{0.5}\qquad or \qquad x>\sqrt{0.5}\\ \text{the distance from $\sqrt{0.5}$ to 1 = $1-\sqrt{0.5} $ }\\ \text{the distance from -1 to $-\sqrt{0.5}$ also = $1-\sqrt{0.5} $ }\\~\\ \text{So the given probability will be }\\ \frac{2(1-\sqrt{0.5})}{1--1}=\frac{2(1-\sqrt{0.5})}{2}=1-\sqrt{0.5}\approx 0.29 \)
