How I can apply conditional probability, formally, to this problem? A fair coin is tossed three times.
What is the probability of two or more heads given that there was at least one head?
I know, I could list out all the sample space and do some elementary probability and determine a straightforward answer. However, for all intents and purposes, I am trying to do this formally and that is where my intuition is breaking down. I really could use some help!
I would like to understand what is incorrect in my approach:
Let $A$ represent the event of two or more heads, and let $B$ represent at least one head. Respectively, we let $A_1$ represent the disjoint event of two heads and $A_3$ represent the event of three heads. Similarily for $B_1, B_2,$ and $B_3$. 
We are trying to show that $P((A_1|B)\cup(A_2|B))$. That is, the probability of two heads given at least one head or the probability of three heads given at least one head.
This, by inclusion-exclusion principle is equal to $P(A_1|B) + P(A_2|B) - P((A_1|B)\cap(A_2|B))$.
However, just by looking at this now, I feel daunted that I am over complicating this problem completely. 
Can anyone help me out here? I want to approach this problem with more than just intuition. 
 A: I personally don't think what you're doing is clear and I think you have over complicated this. For example, it's not clear what $A_2$ is and I don't understand what "disjoint event of two heads" means.

What is the probability of two or more heads given that there was at least one head?

We can use one variable, for example $N$. Let $N$ be the number of heads in 3 tosses.
Then the question becomes $P(N\geq 2| N\geq 1)$. Starting with the definition, we have
\begin{align*}
P(N\geq 2|N\geq 1) &= \frac{P(N\geq 2\cap N\geq 1)}{P(N\geq 1)}\\
&=\frac{P(N\geq 2)}{1-P(N=0)}\tag 1\\
&=\frac{P(N = 2)+P(N=3)}{1-P(N=0)}\tag 2\\
&=\frac{\binom{3}{2}\left(\frac{1}{2}\right)^{2}\left(\frac{1}{2}\right)^{3-2}+\binom{3}{3}\left(\frac{1}{2}\right)^{3}\left(\frac{1}{2}\right)^{3-3}}{1-\binom{3}{0}\left(\frac{1}{2}\right)^{0}\left(\frac{1}{2}\right)^{3-0}}\\
&=\frac{4}{7}
\end{align*}
where I noticed that $N$ follows a binomial distribution with $n = 3, p =1/2$, in $(1)$ the intersection of $N\geq 1$ and $N\geq 2$ is $N\geq 2$, and in $(2)$, \begin{align*}
P(N\geq 2) &= P(N = 2 \cup N=3)\\
&= P(N = 2)+P(N=3)-P(N=2\cap N = 3) \\
&= P(N = 2)+P(N=3)
\end{align*}
since $P(N=2\cap N=3) = P(\varnothing)=0$ because $N =2$ and $N = 3$ are disjoint. Also, I used the complement to find $P(N\geq 1)$ in $(1)$.
