 Solution

Under independence of random variables, the law of a random variable $Z$ is the mixed probability distribution $(1-p){\delta }_0+p\mathrm{Unif}(-1,1)$. For negative $z$, only the uniform component can contribute to $\mathbb{P}(Z\le z)$; at zero the probability mass $1-p$ is included. Thus its cumulative distribution function isThe sketch consists of a horizontal segment at height zero before $-1$, a line of slope $p/2$ ending at the open point $(0,p/2)$, a jump of size $1-p$ to the closed point $(0,1-p/2)$, another line of slope $p/2$ ending at $(1,1)$, and a horizontal segment thereafter. The cumulative distribution function is right-continuous at the jump.

Under the literal assumptions, write $H(z)=\mathbb{P}(U\le z\mid B=1)$. The general answer is $F_Z(z)=(1-p)1_{\{z\ge 0\}}+pH(z)$, so the continuous part of the sketch is undetermined. The jump at zero is still exactly $1-p$, because $\mathbb{P}(U=0)=0$. In the dependent counterexample from (ii), $F_Z(z)=0$ for $z<0$, $F_Z(z)=(1+z)/2$ for $0\le z<1$, and $F_Z(z)=1$ for $z\ge 1$.