An article manufactured by a company consists of three components A, B, and C. These components are manufactured by three independent processes. The probabilities of those components being defective in the process of manufacture are 0.04, 0.03, 0.05 and respectively. What is the probability of the assembled article being defective?
I have already solved this question using tree diagram. $$ Ans = [1- (1-0.04)(1-0.03)(1-0.05) ] $$
But I'm interested to know if I can solve this using total probability theorem.
$$\mathbb{P}(B)=\mathbb{P}(A_1) \mathbb{P}(B\mid A_1)+\mathbb{P}(A_2) \mathbb{P}(B\mid A_2)..$$
Not really, I think. I'll use $A, B, C$ for the event that the respective component is defective. You want $P(\text{defective}) = P(A \cup B \cup C)$ which equals (by the inclusion exclusion formula):
$$P(A) + P(B) + P(C) - P(A \cap B) - P(A \cap C) - P(B \cap C) + P(A \cap B \cap C)$$
and then compute all these using products (as we have independence); not really appealing, but doable.
We can also note that $$P(A \cup B \cup C) = 1- P(A \cup B \cup C)^c) = 1- P(A^c \cap B^c \cap C^c) = 1-P(A^c)P(B^c)P(C^c)$$ using, de complement law, the de Morgan's law and independence of $A,B,C$ in that order; this corresponds to your answer. It's also the fastest one, I think.
To use the law of total probability, we need to find a natural partition $A_1,A_2,A_3, \ldots$ of the set of outcomes, such that $P(\text{defective}|A_i)$ is easily computable and the $P(A_i)$ as well.
One such partition could be $A_1 = A$, $A_2= A^c \cap B$, $A_3 = A^c \cap B^c \cap C$ and $A_4 = A^c \cap B^c \cap C^c$. Then $P(\text{defective}|A_i) = 1$ for $i=1,2,3$ while $P(\text{defective}| A_4) = 0$.
So plugging in, we get another way to get your answer; this is essentially a tree method too.