# Find probability the remaining coin is biased???

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:)

## 1 Answer

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?
• Are you suggesting to simply use the complement? Then P(third is biased) = 1 - P(first two are fair)? – zweifel Sep 21 '20 at 20:50
• whoops meant P(third is biased) = 1-P(first is biased)-P(second is biased) – zweifel Sep 21 '20 at 21:32