If we let $X$ be a random variable with c.d.f. $F(X)=x^3$ for $0 \le x \le 1$ as my cumulative distribution function and I need to find $P(X\ge \frac12)$ is this correct? $$\int_{1/2}^1 x^3dx=\frac{15}{64} $$ or is the correct way to find the probability this way: $$1-\int_0^{\frac12} x^3dx=\frac{64}{65}$$ I think it is the second way because by looking at a graph it would make sense, however, it seems that this probability is too high for it to be correct.
Is this the right probability for this CDF?
0
$\begingroup$
probability
probability-theory