Marginal distribution of a survival function with binary covariate To get the marginal expression of a survival function that includes a binary covariate $Z$ of values 0 and 1, and a covariate rate of 50% (50% of individuals have $Z=1$ and 50% of individuals have $Z=0$), is it correct to simply do:
$$\begin{align}  S(.) &=    \text{covariate rate} \times S(Z=1) +  (1- \text{covariate rate}) \times S(Z=0) \\ &= 0.5 \times S(Z=1) +  0.5 \times S(Z=0) \end{align}  $$
Edit
I thought that $Z$ followed a binomial distribution and that I should use the probability mass functions such as:
$$\Pr(Z = k) = \binom{n}{k}p^k(1-p)^{n-k}$$
Thanks to @winperikle's previsions, it turns out that Z actually follows a Bernoulli distribution.
Thanks in advance for your answers!
 A: If the survival function of a random variable $X$ is defined conditionally to a variable $Z$, 
$$
S(t \mid Z) = \mathbb P (X \geq t \mid Z)
$$
then the marginal survival is
$$
S(t) = \mathbb E_Z[ S(t \mid Z) ] = \mathbb E_Z[ \mathbb P (X \geq t \mid Z) ].
$$
From the law of the unconscious statistician we have,
\begin{align*}
\mathbb E_Z[ S(t \mid Z) ] &= \sum_z S(t \mid z) \mathbb P(Z=z) \\ 
&= S(t \mid z=0) \mathbb P(Z=0) + S(t \mid z=1)\mathbb P(Z=1) \quad \text{(if $Z$ is binary)}.
\end{align*}
In your case you have $\mathbb P(Z=0) = \mathbb P(Z=1) = 0.5$.

The binary variable $Z$ follows a Bernoulli distribution.
A Binomial distribution is defined as the sum of several independent Bernoulli-distributed random variables. So a Bernoulli can be seen as Binomial with only one variable in the sum.
Then, if $n=1$ we have,
$$
\binom{1}{k}p^k(1-p)^{1-k} = 1-p \quad \text{if $k=0$}
$$
and 
$$
\binom{1}{k}p^k(1-p)^{1-k} = p \quad \text{if $k=1$}
$$

Example 
If we have $X \geq 0$, define for $t\geq 0$, 
$$
\mathbb P( X \geq t \mid Z) = e^{-t}I_{Z=0} + e^{-2t}I_{Z=1}
$$
Then the marginal survival is simply, if $\mathbb P(Z=1) = \frac{3}{4}$,
$$
\mathbb P(X \geq t ) = \frac{e^{-t}}{4} + \frac{3 e^{-2t}}{4}.
$$

I hope this answers your question.
