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I've been given an interesting homework question:

Given three coins in a box, where two have a probability of heads ( $p(h) = 0.5$ ), and the third has $p(h) = \frac{1}{3}$. I draw one coin, and the result is a tail. What is the probability that I did not remove the unfair coin from the box?

I started solving this by trying to find $p(\text{unfair}\mid t)$, and constructed a sample space. Let $A,B$ represent the fair coins, and $C$ represent the unfair one. The sample space is:

$$ \{A_t, A_h, B_t, B_h, C_t, C_h \} $$

With probabilities of

$$ \left\{ \frac{1}{6},\frac{1}{6},\frac{1}{6},\frac{1}{6},\frac{2}{9},\frac{1}{9} \right\} $$

In trying to find $p(\text{unfair}\mid t)$ we reduce this sample space to:

$$ \{A_t, B_t, C_t \} $$

I can't figure out what the new probabilities for this sample space are though. I think $C_t$ has a higher probability than the other two, but I can't seem to divide this into three probabilities that add to 1.

How would I go about finding the new probabilities of this subspace? I've tried a bunch of tricks such as multiplying the old probabilities by 2, but then the sample space has a total probability > 1.

Any help is appreciated, tips included. I would like to be able to do this myself so don't give me the answer outright, please.

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    $\begingroup$ What is the probability that you get $T$ from the toss? What portion of that probability is explained by the scenario in which you chose the biased coin? $\endgroup$ – lulu Jun 16 '19 at 15:12
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    $\begingroup$ Do you know Bayes Theorem? Because this question is basically asking you to construct its reasoning. $\endgroup$ – Alexander Geldhof Jun 16 '19 at 15:14
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    $\begingroup$ Your trick with the multiplication by $2$ is the right idea in this particular case. However, as you note, the sum of the probabilities is then larger than $1$. However, have you tried multiplying by the number which $would$ make the sum of the probabilities $1$? $\endgroup$ – Alexander Geldhof Jun 16 '19 at 15:16
  • $\begingroup$ @AlexanderGeldhof Yes I know Bayes Theorem, I find this to easier to understand and more intuitive when solving an unfamiliar problem though. Your second comment makes an unreasonable amount of sense now that I look at it - thanks. If you posted an answer saying that I would accept it ;). $\endgroup$ – Howard P Jun 16 '19 at 15:19
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    $\begingroup$ Right, so can you compute the probability of getting $T$ on that first toss? $\endgroup$ – lulu Jun 16 '19 at 15:24
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Your trick with the multiplication by $2$ is the right idea in this particular case. However, as you note, the sum of the probabilities is then larger than $1$. However, have you tried multiplying by the number which would make the sum of the probabilities $1$?

(Note that this argument relies on the fact that the chances for drawing each coin from the jar are equal)

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Probability of getting a tail by drawing a coin assumming it is equally probable to draw a coin from any jar.

$P(T) = \frac{1}{3}\frac{1}{2}+\frac{1}{3}\frac{1}{2}+\frac{1}{3}\frac{2}{3} = \frac{5}{9}$

Probability of getting a tail by drawing fair coins $= \frac{1}{3}$

Thus Proababiltiy that you did not pick the unfair coin $= \dfrac{\frac{1}{3}}{\frac{5}{9}} = \frac{3}{5}$

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