We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive pseudonymised information about your use of our website. cookie policy and privacy policy.
 
+0  
 
0
62
1
avatar

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
 #1
avatar+103693 
+2

If 

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

.
 Aug 12, 2019

28 Online Users

avatar
avatar
avatar
avatar
avatar