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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.

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  • $\begingroup$ Welcome to MathSE. Please read this tutorial on how to typeset mathematics on this site. $\endgroup$ – N. F. Taussig Sep 4 '18 at 8:19
  • $\begingroup$ @N.F.Taussig Hi, I was struggling typsetting in Latex or MathJax fractions with words in it. Should i use \mbox? $\endgroup$ – bob bob Sep 4 '18 at 8:24
  • $\begingroup$ A simpler solution is possible. Let $A$ be the event of getting at least one heads, and $B$ the event of getting at least two heads. Notice that $A \cap B = B$. It follows that $P(A|B) = P(A\cap B)/P(B) = P(B)/P(B)=1$. $\endgroup$ – littleO Sep 4 '18 at 8:45
  • $\begingroup$ I have edited your equations using the align environment. I introduced the intersection symbol, which can be typeset using \cap when you are in math mode. I also moved the text inside the fractions. See if you are happy with the results. If you right-click on the equations, you will see a box that says Show Math As at the top. If you click this, you can select the option TeX commands. Alternatively, you can click on the edit button to see what I did or improve on it. $\endgroup$ – N. F. Taussig Sep 4 '18 at 8:48
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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 $3p^2(1-p)$

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  • $\begingroup$ thank you so my answer is correct? $\endgroup$ – bob bob Sep 4 '18 at 8:11
  • $\begingroup$ @bobbob It would have been better if you had simplified your final expression to $1$ $\endgroup$ – Henry Sep 4 '18 at 9:50

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