Calculate P(A|AUBUC) independent events

Question

For three events independent events $A,B,C$ with $P(A)=0,7$, $P(B)=0,4$, $P(C)=0,3$ i want to find $P(A|AUBUC)$. I know that $A,B,C$ independent means $P(ABC)=P(A)P(B)P(C)$ and $P(AB)=P(A)P(B), P(AC)=P(A)P(C), P(BC)=P(B)P(C)$ can i use Bayes $$P(A|AUBUC)=\dfrac{P(AUBUC|A)P(A)}{P(AUBUC)}$$ but how can i calculate P(AUBUC|A) ?

Answer 1

$$P(A \cup B \cup C|A)=1$$

You can evaluate $P(A \cup B \cup C)$ using the following \begin{align} P(A \cup B \cup C) &= 1-P(A^cB^cC^c) \end{align}

and the fact that they are independence.

Answer 2

Note that $A\subseteq A\cup B\cup C$.

In general if $A\subseteq D$ then $P(A\mid D)=P(A\cap D)/P(D)=P(A)/P(D)$

If $A,B,C$ are independent then so are $A^{\complement},B^{\complement},C^{\complement}$ so that: $$P(A\cup B\cup C)=1-P((A\cup B\cup C)^{\complement})=1-P(A^{\complement}\cap B^{\complement}\cap C^{\complement})=1-P(A^{\complement})P(B^{\complement})P(C^{\complement})=1-(1-P(A))(1-P(B))(1-P(C))$$

Answer 3

Alternatively: $$P(A\cup B\cup C)=P(A)+P(B)+P(C)-P(A\cap B)-P(A\cap C)-P(B\cap C)+P(A\cap B\cap C)=0.7+0.4+0.3-0.28-0.21-0.12+0.084=0.874.$$ Hence: $$P(A|A\cup B\cup C)=\frac{P(A\cap (A\cup B\cup C))}{P(A\cup B\cup C)}=\frac{P(A)}{P(A\cup B\cup C)}=\frac{0.7}{0.874}\approx0.8.$$