Conditional probability (two coins) Two coins are flipped, if at least one them lands heads up, what is the probability that both do?
I've tried solving this problem countless different ways but I can never get it right.
I know that "at least one" means "one or all" so I use P(AorB) then the result I divide it over the total number of ways the event can occur.
 A: Intuitive Answer:
There are $3$ options:


*

*$HH$

*$HT$

*$TH$


Only $1$ of these options is $HH$, hence the probability is $\frac13$.

Formal Answer:
Let $A$ denote the event in which exactly two of the coins landing on $H$.
Let $B$ denote the event in which at least one of the coins landing on $H$.
The probability that event $A$ has happened given that event $B$ has happened:
$$P(A|B)=\frac{P(A\cap B)}{P(B)}=\frac{1/4}{3/4}=\frac13$$
A: Let $A=$Both are head, $B=$At least one is head. 
Then you are asked to find $\displaystyle P(A/B)=\frac{P(A\cap B)}{P(B)}$.
Sample space=$\{HH,HT,TH,TT\}$. Then $B=\{HT,TH,HH\}$. So $P(B)=3/4$. 
$A\cap B=\{HH\}$, so $P(A\cap B)=1/4$. So,$\displaystyle P(A/B)=1/3$.
A: 
I've tried solving this problem countless different ways but I can never get it right. I know that "at least one" means "one or all" so I use P(AorB) then the result I divide it over the total number of ways the event can occur.

Not quite so.   You are after $\mathsf P(A\cap B\mid A\cup B)$, where $A,B$ are the events of obtaining heads on each coin, respectively.    Since $(A\cap B)\subset (A\cup B)$ just use the definition of condition.
$$\mathsf P(A\cap B\mid A\cup B) = \dfrac{\mathsf P(A\cap B)}{\mathsf P(A\cup B)}$$

Remark: $\cap$ is the set intersection ("and") operator, and $\cup$ is the set union ("or") operator.
$\mathsf P(A\cap B)$ is the probability that both coins are heads.
$\mathsf P(A\cup B)$ is the probability that at least one is.
$\mathsf P(A\cap B\mid A\cup B)$ is the probability that both coins are heads when given that at least one is.
