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There are 2 coins, one fair (heads/tails) and one unfair (head/head)
A person chooses randomly between the two coins, and throws it. the coin lands on head.
he throws the same coin again and it lands on head again.
he throws the same coin the third time and get tails

the question is what is the probability he chose the fair coin.
now, it sounds dumb as there is no tails to the unfair coin so the probability of him choosing the fair coin while twice landing it on head and third time is on tail is $1$

am I right? what is the math equation for this? I tried using bayes' law however I got confused with the 3 different throws.. Thanks!

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  • $\begingroup$ Let $\textrm{hht}$ be the event that a coin, $c_1$ (fair) or $c_2$, shows two times head and one time tail. Then you are asked to calculate $P(c_1|\textrm{hht})=\frac{P(c_1 \cap \textrm{hht})}{P(\textrm{hht})}$. What´s the problem with that? $\endgroup$ – callculus Apr 25 at 20:32
  • $\begingroup$ @callculus I don't know, is it 1 ? what is the formula for that i dont understand, how do you calculate $P(c_1 \cap hht)$ whatever $hht$ is... $\endgroup$ – SendHelp Apr 25 at 20:38
  • $\begingroup$ hht is the event that the outcome is head, head, tail. So you calculate the probability that the fair coin has been chosen and the outcome is hht. $\endgroup$ – callculus Apr 25 at 20:42
  • $\begingroup$ @callculus is it 1 ? $\endgroup$ – SendHelp Apr 25 at 20:44
  • $\begingroup$ I don't know, is it 1 ? I don´t know either in advance. Just go the systematic way. Math has nothing to do with guesswork. $\endgroup$ – callculus Apr 25 at 20:44
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The sample space of any outcome that results in tails for the unfair coin is 0, therefore it doesn't matter how many heads you toss in a row, if tails ever appears, then the probability that the coin you chose is the fair one is 1.0.

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