Conditional probability question with three independent events I have the following probabilities:
$$P(A_1)= P(A_2) = P(A_3)= .2,$$
where $A_1, A_2,$ and $A_3$ are all independent.
I wish to find
$$P(A_1\mid A_1 \cup A_2 \cup A_3).$$
I know that
$$P(A_1 \cup A_2 \cup A_3) = 3(.2) - 3(.2)(.2) + (.2)(.2)(.2) = .488.$$
I do not know how to proceed after this, however.  I do not understand how to incorporate the conditional part of the problem into my answer.
Any guidance is much appreciated.  Thanks!
Just use the definition of Conditional Probability, and the fact that $A_1\subseteq A_1\cup A_2\cup A_3$.
$$\mathsf P(A_1\mid A_1\cup A_2\cup A_3)=\dfrac{\mathsf P(A_1)}{\mathsf P(A_1\cup A_2\cup A_3)}=\dfrac{0.200}{0.488}$$
Note that $$\mathbb{P}(A_{1}|A_{1}\cup A_{2}\cup A_{3})=\frac{\mathbb{P}(A_{1}\cap(A_{1}\cup A_{2}\cup A_{3}))}{\mathbb{P}(A_{1}\cup A_{2}\cup A_{3})}\quad (\text{by definition}).$$Now, since you know that $\mathbb{P}(A_{1}\cup A_{2} \cup A_{2})=0.488$ and $A_{1}\cap (A_{2}\cup A_{2} \cup A_{3})=A_{1}$, so you have $$\mathbb{P}(A_{1}|A_{1}\cup A_{2}\cup A_{3})=\frac{0.200}{0.800}.$$
EDIT: We have that $$\boxed{\mathbb{P}(A_{1}|A_{1}\cup A_{2}\cup A_{3})=\frac{0.200}{0.488}}.$$
Also you can see that $$\mathbb{P}(A_{1}\cup A_{2} \cup A_{3})=\mathbb{P}(A_{1})+\mathbb{P}(A_{2})+\mathbb{P}(A_{3})-\mathbb{P}(A_{1}\cap A_{2})-\mathbb{P}(A_{1} \cap A_{3})-\mathbb{P}(A_{2}-A_{3})+\mathbb{P}(A_{1}\cap A_{2} \cap A_{3})$$
since $A_{1},A_{2},A_{3}$ are three independent events, so $$\mathbb{P}(A_{1} \cup A_{2} \cup A_{3})=3(0.2)-3(0.2)^{2}+(0.2)^{3}=0.488.$$