= Solution
Under <independence of random variables>, the <law of a random variable> $Z$ is the <mixed probability distribution> $(1-p)\delta_0+p\operatorname{Unif}(-1,1)$. For negative $z$, only the uniform component can contribute to $\mathbb P(Z\leq z)$; at zero the <probability mass> $1-p$ is included. Thus its <cumulative distribution function> is
$$
\boxed{F_Z(z)=\begin{cases}0,&z<-1,\\p(z+1)/2,&-1\leq z<0,\\1-p+p(z+1)/2,&0\leq z<1,\\1,&z\geq1.\end{cases}}
$$
The 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.
\Image[/past-exam-of-the-probability-course-of-the-university-of-oxford/2023-bernoulli-uniform-cdf.png]
{title=Cumulative distribution function of the independent Bernoulli-uniform product, illustrated at $p=0.4$}
Under the literal assumptions, write $H(z)=\mathbb P(U\leq z\mid B=1)$. The general answer is $F_Z(z)=(1-p)\mathbf1_{\{z\geq0\}}+pH(z)$, 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 $0\leq z<1$, and $F_Z(z)=1$ for $z\geq1$.
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