One tails from a mostly-fair drawer of coins

Question

A drawer holds 10 coins: 8 are fair, and 2 are biased with $P(\text{H}) = 0.75$ . You grab one at random, flip it once, and get tails.

What is the probability the coin you grabbed is one of the biased ones?

Accepted Answer

Priors from the drawer: P(	ext{fair}) = 0.8, P(	ext{biased}) = 0.2. Likelihoods of tails: P(	ext{T}\mid 	ext{fair}) = 0.5,\qquad P(	ext{T}\mid 	ext{biased}) = 1 - 0.75 = 0.25. Bayes numerators: 	ext{fair: } 0.8	imes 0.5 = 0.40,\qquad 	ext{biased: } 0.2	imes 0.25 = 0.05. Normalize: P(	ext{biased}\mid 	ext{T}) = \frac{0.05}{0.40 + 0.05} = \frac{0.05}{0.45} = \frac{1}{9} \approx 0.111. A tails result actually makes the biased coin less likely than its 20% prior, because biased coins rarely produce tails. Takeaway. Evidence can push a hypothesis's probability down: tails is weak for t…

One tails from a mostly-fair drawer of coins

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