We place three coins in a box. One coin is two-headed. Another coin is two tailed. Only the third coin is real, but they all look alike. Then we randomly select one coin from this box, and we flip it. If heads comes up, what is the probability that the other side of the coin is also heads?

Let $H$ represent the event that heads comes up on tossing a coin drawn randomly. $ P(H) = \frac{1}{3} + \frac{1}{3} \frac{1}{2} = \frac{1}{2}$. Let $I$ be the event of other side being head. Do I need to compute $P(I|H)?$

$P(H \cap I) = P(\text{choosing the coin with 2 heads}) = \frac{1}{3}$. Then $P(I|H) = \frac{P(H \cap I)}{P(H)} = \frac{2}{3}.$ Is this correct?

  • $\begingroup$ Yes. That is correct. I will mention that you have typeset it as $P(I/H)$, you should instead use a vertical line to denote conditional probability. $P(I|H)$. There was also another small typo which I took the liberty of correcting where you accidentally did $P(H|I)$. $\endgroup$ – JMoravitz Feb 12 '15 at 2:20
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    $\begingroup$ Actually, these coins are either all real or all imaginary ... perhaps you meant to say "only one of them is normal"? :-) $\endgroup$ – Dale M Feb 12 '15 at 4:40
  • $\begingroup$ Also the two-tailed one doesn't look like the two-heading one from any angle except possibly looking at it edge-on so you can't see either face. I think you mean we don't have any way to pick one more likely than the other when we choose randomly. $\endgroup$ – David K Feb 12 '15 at 6:00
  • $\begingroup$ But don't worry about the nit-picking; you did fine. $\endgroup$ – David K Feb 12 '15 at 6:02

Your computation is correct. I have tried to make things a bit more explicit.

Let $H$ be the event that we see heads and $C_2$ we chose the $2$-headed coin, $C_1$ we chose the normal coin, and $C_0$ we chose the $2$-tailed coin. Then

$P(H\cap C_2)=P(C_2)P(H\mid C_2)=\frac13\cdot1=\frac13$
$P(H\cap C_1)=P(C_1)P(H\mid C_1)=\frac13\cdot\frac12=\frac16$
$P(H\cap C_0)=P(C_0)P(H\mid C_0)=\frac13\cdot0=0$

Since $C_0$, $C_1$, and $C_2$ are disjoint and all-inclusive, we have, as expected, that

$P(H)=P(H\cap C_2)+P(H\cap C_1)+P(H\cap C_0)=\frac13+\frac16+0=\frac12$

Since the only way the other side can also be a head is with the $2$-headed coin, what we want is

$P(C_2\mid H)=\frac{P(H\cap C_2)}{P(H)}=\frac{1/3}{1/2}=\frac23$

Another way to look at this is to label the sides of each coin as follows $$ \begin{array}{c} \text{coin}&\text{A}&\text{B}\\ \hline \text{two head}&\text{head}&\text{head}\\ \text{normal}&\text{head}&\text{tail}\\ \text{two tail}&\text{tail}&\text{tail}\\ \end{array} $$ Each of the $6$ possibilities is equally likely. We have chosen one of the three heads, and two of those choices give the flip side as a head. Thus, the probability is $\frac23$.


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