What is the $P(X=0)$ where $X$ is a random variable? Given the probability mass function $f_X(x)=.1\delta(x)+.9u(x)e^{-x}$ find the probability that $P(X=0)$? I am a little bit confused with $\delta$ function. I know that if we have continuous random variable $P(X=a)=0$ for all $a$ values. So the probability in case of the given random variable would be zero. But $\delta(0)$ is 0.1. Can someone entangle this for me?

The $\delta$ is the Dirac delta impulse function. This has the property $\int_\Bbb R \delta(x)\operatorname d x = 1$

Sometimes written as $\displaystyle\lim_{A\to\\{0\\}}\int_A \delta(x)\operatorname d x=1\;,\, \lim_{A\to\Bbb R\setminus\\{0\\}}\int_A \delta(x)\operatorname d x=0$

Similarly the Heaviside step function being used appears to be, $u(x) = \begin{cases} 1 &:& x>0\\\0&:&x\leq 0\end{cases}$

Basically the use of $\delta$ and $u$ indicates that you have a mixture distribution, with a probability mass at a discrete point and a probability density elsewhere in the support. (But where and how much?)

So what you are looking for is:

$$\mathsf P(X=0) =\lim_{A\to\\{0\\}}\int_A 0.1\delta(x)+0.9 u(x)e^{-x}\operatorname d x \color{white}{= 0.1}$$