Given $X$ is uniform random variable, $P\{X>1\} = 0.6$ and $F(2) = 0.5$.
Find $P\{-1\leq X < 3\}$.
My solution is: $P\{X>1\} = F(\infty) - F(1) = 0.6$. So $F(1) = 0.4$
And now I assume that F grows linearly. I need to find $F(3) - F(-1)$. Using fact of linearity I can say that $F(3) - F(-1) = 4 * [F(2) - F(1)]$.
My answer is $4\times(0.5 - 0.4) = 0.4$.
Am I right?
If $X$ is uniform on some support $a \le X \le b$, then the density is $$f_X(x) = \begin{cases} \frac{1}{b-a}, & a \le x \le b, \\ 0, & \text{otherwise}. \end{cases}$$ The CDF is then $$F_X(x) = \begin{cases} 0, & x < a \\ \frac{x-a}{b-a}, & a \le x \le b \\ 1, & b < x. \end{cases}$$ We are given $$F_X(1) = 1 - \Pr[X > 1] = 1 - 0.6 = 0.4,$$ and $$F_X(2) = 0.5.$$ From these, we determine $$\frac{1 - a}{b-a} = 0.4, \\ \frac{2 - a}{b-a} = 0.5.$$ Hence $$\frac{0.4}{0.5} = \frac{1-a}{2-a},$$ so $a = -3$; we then substitute back to obtain $b = 7$, and $$\Pr[-1 \le X < 3] = F_X(3) - F_X(-1) = \frac{3-(-3)}{7-(-3)} - \frac{-1-(-3)}{7-(-3)} = \frac{2}{5}.$$
The reason why your solution works is that it so happens that the given probabilities result in the support $[a,b]$ to be "large enough" to contain the desired outcome $-3 \le X < 1$. For example, now that we know $[a,b] = [-3,7]$, if I asked you for $\Pr[-5 < X < 0]$, you could neither say this is $1$ nor $0.5$. The answer is $0.3$ because $\Pr[-5 < X < 0] = \Pr[-3 < X < 0]$, since the intersection $$(-5,0) \cap [-3,7] = [-3, 0).$$ So you need an extra step to justify that the "shortcut" you used is valid.