= Solution
{c}
For the <exponential distribution> with rate $\lambda$, the <cumulative distribution function> is $F(x)=1-e^{-\lambda x}$ for $x\geq0$, and zero for $x<0$. Its <quantile function> is
$$
\boxed{F^{-1}(u)=-\frac{\log(1-u)}{\lambda},\qquad0<u<1.}
$$
For <inverse transform sampling>, draw $U$ from the <uniform distribution> on $(0,1)$ and return $-\log(1-U)/\lambda$. Equivalently return $-\log U/\lambda$, since $1-U$ has the same <uniform distribution>. Indeed, for $x\geq0$, $\mathbb P(-\log U/\lambda>x)=\mathbb P(U<e^{-\lambda x})=e^{-\lambda x}$.
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