Suppose that we have $2$ coins, a fair one (probability of $H$ or $T$ = $0.5$), lets call this $C_1$ and a biased one (probability of $H = 0.6,$ $T = 0.4$), lets call this $C_2$.

Then if we pick one of the coins at random and look at the result of a coin toss for which we get a $H$ result, why can we say that the probability of the coin being $C_1 = 0.4$ and being $C_2 = 0.6$ respectively.


1 Answer 1


Use Bayes' Theorem

$$\begin{align*} P(\text{Coin 1 | } H) &=\frac{P(\text{Coin 1} \cap H)}{P(H)}\\\\ &=\frac{0.5\cdot0.5}{(0.5\cdot0.5)+(0.5\cdot0.6)}\\\\ &\approx0.455 \end{align*}$$

and so $$P(\text{Coin 2 | } H)=1-P(\text{Coin 1 | } H)\approx0.545$$

It does not appear that the probability that it was coin one given that you obtained heads it $0.4$.

  • $\begingroup$ Thanks for your answer. My question was based on a question posted on stack exchange: cs.stackexchange.com/questions/76647/… $\endgroup$ Commented Mar 1, 2018 at 20:27
  • $\begingroup$ I could not understand where the values were coming from $\endgroup$ Commented Mar 1, 2018 at 20:28
  • 1
    $\begingroup$ As someone in the comments pointed out, there were errors in the accepted answer. $\endgroup$
    – Remy
    Commented Mar 1, 2018 at 20:30

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