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My try :

we can deal three different cases of removing two biased coin and 1 unbiased coin.

Individual probability is known $$\frac9{10} , \frac2{25},\frac1{50}$$

After that total is 49 as one is removed. now what to do conditional probability or Bayes' theorem ? i dont know how to implement ?

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There are four possible combinations:

  • Double-headed and shows heads with probability $\frac{4}{50}\times 1= 0.08$
  • Double-tailed and shows tails with probability $\frac{1}{50}\times 1 =0.02 $
  • Fair coin and shows heads with probability $\frac{45}{50}\times\frac12 = 0.45$
  • Fair coin and shows tails with probability $\frac{45}{50}\times\frac12 = 0.45$

so using ${P(A,B)}=P(A \mid B) \,P(B)$, i.e. $P(A \mid B) =\dfrac{P(A,B)}{P(B)}$, and $P(B)= P(A,B)+P(A^c,B)$, you get

(a) Given it shows heads, the conditional probability of being double-headed is $\frac{0.08}{0.08+0.45}$

(b) Given it shows heads, the conditional probability of being fair is $\frac{0.45}{0.08+0.45}$

(c) Given it shows tails, the conditional probability of being fair is $\frac{0.45}{0.02+0.45}$

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Here's how I think about these problems. There are $50$ coins with $2$ sides each, or $100$ possible outcomes, each equally likely. Of those possible outcomes, $53$ result in heads and $47$ result in tails.

Out of the $53$ possibilities that result in heads, $8$ of them come from a two-headed coin and the other $45$ come from fair coins. So if your result is heads, the probability of a two-headed coin is $\frac{8}{53}$ and the probability of a fair coin is $\frac{45}{53}.$ If the result is tails, the probability of a fair coin is $\frac{45}{47}$.

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  • $\begingroup$ +1 Indeed it is very handsome to look at sides instead of coins in problems like this. $\endgroup$ – drhab Jan 20 at 9:28

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