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Here is the layout of the question:

3 coins, 2 are fair but the third is biased with a probability $4/7$ for heads. You randomly choose one of these coins, tossing it 10 times, and obtain 5 heads 5 tails. You then choose a different coin, again toss 10 times, and obtain 4 heads 6 tails. What is the probability that the remaining coin is the biased coin?

I have a feeling Bayes theorem could be helpful with this problem but I am unsure how and where to apply it. Is this just the multiplication of two conditional probabilities (namely $P(\text{first is fair}|5H,5T)$ and $P(\text{first is fair}|4H,6T)$)? Thanks:)

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Perhaps, first try to answer these:

  • If the first coin is biased, what is the probaility of the observed outcomes for the two coins?
  • If the second coin is biased, what is the probaility of the observed outcomes for the two coins?
  • If the third coin is biased, what is the probaility of the observed outcomes for the two coins?
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  • $\begingroup$ Are you suggesting to simply use the complement? Then P(third is biased) = 1 - P(first two are fair)? $\endgroup$
    – zweifel
    Sep 21, 2020 at 20:50
  • $\begingroup$ whoops meant P(third is biased) = 1-P(first is biased)-P(second is biased) $\endgroup$
    – zweifel
    Sep 21, 2020 at 21:32

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