Rewrite the probability statement $P(0<R<k(X_1+X_2))=1-S$

Question

Rewrite the probability statement $$P(0<R<k(X_1+X_2))=1-S$$ into the form $$P(X_1+X_2<f(R,k))=S$$ where $X_1$ and $X_2$ constitute a random sample from a uniform population with parameters $0$ and $R$, and $f(R,k)$ is a function of $R$ and $k$.

So far, I've figured out that $P(R<0)+P(R>k(X_1+X_2)=S$ and so $P(k(X_1+X_2)<R)+P(R<0)=S$ $\implies$$P(X_1+X_2<\frac1kR)+P(R<0)=S$

I'm having trouble dealing with the $P(R<0)$ part because I somehow need to combine the sum of these probabilities all into one statement, but I'm not sure how.

Answer 1

Assuming $k>0$

$P(0<R<k(X_1+X_2))=P(X_1+X_2>R/k>0)=1-S\Rightarrow 1-P(X_1+X_2>R/k>0)=S\Rightarrow P(X_1+X_2<R/k)=S\space \text{if $R>0$}\space, \text{otherwise}\space S=0$