Originally Posted by

**SlipEternal** The first probability is 1, as the maximum value for X is 0.5. The second probability will be 0.5. The third probability is 0. The last probability is 0.

The first probability is $P([0,1]) = 1-0 = 1$.

The second probability is $P\left( \left[ 0, \dfrac{1}{4} \right] \cup \left[ \dfrac{3}{4}, 1 \right] \right) = P\left( \left[ 0, \dfrac{1}{4} \right] \right) + P\left( \left[ \dfrac{3}{4}, 1 \right] \right) = \dfrac{1}{4} - 0 + 1 - \dfrac{3}{4} = \dfrac{1}{2}$

The third probability is $P(\{0,1\}) = P([0,0]) + P([1,1]) = 0-0+1-1 = 0$

The last probability is $P(\emptyset) = 0$.