Probability function of three Random Variables multiplied, solidifying intuition

Question

I ran across this exercise:

Let $T$ be a random variable distributed as a $\text{Bernoulli}(p)$, $U$ be a random variable distributed as a $\text{Bernoulli}(q)$ and $W$ be a random variable distributed as a $\text{Poisson}(\lambda)$. Find the probability function of $X =T \cdot U\cdot W\,$ .

I first calculated:

$ \ \ P(T\cdot U = 1) = P(T=1)\cdot P(U=1) = pq \\ P(T \cdot U = 0) = 1 - P(T\cdot U=1) = 1 - pq $

Then I made a mistake thinking that:

$ P(X=0) =P(T\cdot U = 0 , W> 0) +P(T \cdot U = 1, W= 0) $

was the correct formula for the probability of $ X = 0$, this seemed reasonable and symmetrical to me, my intuition lead me astray.

I then thought that the $ P(X = 0) = P(T\cdot U= 0) + P(W = 0)$

For the second time, my intuition lead me astray.

The solutions highlight as the correct solution : $P(X= 0) = P(T\cdot U = 0) + P(T\cdot U = 1, W = 0) $

Intuitively I explained this result to myself as: if $ P(T\cdot U = 0) $ then I do not need to check the other random variable because anything times 0 is 0. If $ P(T\cdot U = 1) $ then I need the Poisson to be 0 to have $ P(X = 0) $.

But because my intuition failed me I would like to understand which probability law the solution is using to solve this exercise, It seems to me not the Total Probability law or any form of Bayes rule. A Formal proof of the equality being used would be helpful to me. Also If you want to point out in witch way my first two attempts where wrong I would be grateful.

Answer 1

It appears that the three variables are independent, although the OP does not explicitly mentions it anywhere.

One does not need to bring in Laws and Theorems involving conditional probabilities here. If the OP had listed from the beginning all possible joint events, he would very intuitively arrived at the correct solution:

$$\begin{align}\\ \{T\cdot U=0, W=0\} & \{T\cdot U=0, W>0\} \\ \{T\cdot U=1, W=0\} & \{T\cdot U=1, W>0\} \\ \end{align}$$

These are mutually exclusive events and they partition the space where $X$ lives. So their probabilities can be added. Looking at the above we can see that the solution provided is correct, and also that:

The OP's first attempt

$$P(X=0) = P(T\cdot U = 0 , W> 0) +P(T \cdot U = 1, W= 0) \text{(false)}$$

misses the probability of $\{T\cdot U = 0 , W= 0\}$.

The OP's second attempt

$$P(X = 0) = P(T\cdot U= 0) + P(W = 0) \text{(false)}$$ double-counts the probability of $\{T\cdot U = 0 , W= 0\}$, since

$$P(T\cdot U= 0) = P(T\cdot U=0, W=0) + P(T\cdot U=0, W>0)$$ while $$P(W = 0) = P(T\cdot U=0, W=0) + P(T\cdot U=1, W=0)$$

In both cases therefore the OP "mistreated" the event $\{T\cdot U=0, W=0\}$ -and this is probably because he thought in terms "$X$ will be zero if $T\cdot U=0$ or $W=0$, and ignored the case were both are zero.