A coin with probability, $p$, of coming up heads is flipped three times (the flips are independent). What is the probability of getting at least one head given there are at least two heads?
My attempt at a solution: First I realize that there are eight possible outcomes: TTT, HHH, THH, HHT, HTH, TTH, THT, HTT. Now consider: \begin{align*} &P(1~\text{head} \mid \text{at least $2$ heads}) + P(2~\text{heads} \mid \text{at least $2$ heads})+ P(3~\text{heads} \mid \text{at least $2$ heads})\\ & \quad = \frac{P(1~\text{head}) \cap P(\text{at least $2$ heads})}{P(\text{at least $2$ heads})} + \frac{P(2~\text{heads}) \cap P(\text{at least $2$ heads})}{P(\text{at least $2$ heads})}\\ & \qquad + \frac{P(3~\text{heads}) \cap P(\text{at least $2$ heads})}{P(\text{at least $2$ heads})}\\ & \quad = 0 + \frac{P(2~\text{heads})}{P(\text{at least $2$ heads})} + \frac{P(3~\text{heads})}{P(\text{at least $2$ heads})}\\ & \quad = \frac{3p^2(1-p)}{3p^2(1-p)+ p^3} + \frac{p^3}{3p^2(1-p)+ p^3} \end{align*}
Is this solution correct? I'm not sure whether the probability of getting two heads is $3p^2(1-p)$ or if it is $(p^2)(1-p)$. I figured that since there was three possible ways of getting two heads I would multiply by $3$ (add three times). But I'm not sure if this correct.
If you have at least two heads then you definitely have at least one head. So you would have thought a probability of $1$ and indeed:
$$0+\frac{3p^2(1-p)}{3p^2(1-p)+ p^3}+ \frac{p^3}{3p^2(1-p)+ p^3} = 1$$
The probability of exactly $k$ heads from $n$ attempts is ${n \choose k}p^k(1-p)^{n-k}$
which with $n=3$ and $k=2$ is equal to $3p^2(1-p)$