What is the relationship between $P(\theta a + (1-\theta)b > 1)$ and $P(a > 1)$, $P(b > 1)$?

Question

Suppose random value $c$ is a mixture of $a$ and $b$ $$c = \theta a + (1-\theta)b$$ What is the relationship of $P(c > 1)$ and $P(a>1)$, $P(b>1)$, is it the following equation? $$ P(c > 1) = \theta P(a>1) + (1-\theta)P(b>1) $$

Answer 1

Unfortunately no, as addition corresponds to convolution in the space of the PDF/CDF. Specifically if $a,b$ are independent:

$P(c>1) = P(\theta a+(1-\theta)b>1)=\int_{-\infty}^\infty P(\theta a> t) P((1-\theta)b>1-t)dt $

As a counterexample to your statement, suppose that $P(a=100)=1$ and $P(b=0)=1$ (constants). Then $P(c>1)=1$ when $\theta>1/100$, but you would have guessed it's $\theta\cdot 1 + (1-\theta)\cdot0=\theta$,