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Let x be a real number randomly chosen so that $$-1 \le x \le 1$$. What is the probability that $$x^2 > \frac12$$?

Aug 12, 2019

$$x^2>0.5\\ then\\ x<-\sqrt{0.5}\qquad or \qquad x>\sqrt{0.5}\\ \text{the distance from \sqrt{0.5} to 1 = 1-\sqrt{0.5}  }\\ \text{the distance from -1 to -\sqrt{0.5} also = 1-\sqrt{0.5}  }\\~\\ \text{So the given probability will be }\\ \frac{2(1-\sqrt{0.5})}{1--1}=\frac{2(1-\sqrt{0.5})}{2}=1-\sqrt{0.5}\approx 0.29$$