In a set of $10$ coins, $2$ coins are with heads on both the sides. A coin is selected at random from this set and tossed five times. If all the five times, the result was heads, find the probability that the selected coin had heads on both the sides.

Using Bayes' theorem, let E1 be choosing a double-headed coin and E2 be choosing a fair coin, with prior probabilities P(E1) = 2/10 and P(E2) = 8/10. The probability of getting five heads with a double-headed coin is 1, and with a fair coin it is (1/2)^5 = 1/32. The posterior probability is [ (2/10) * 1 ] / [ (2/10) * 1 + (8/10) * (1/32) ], which simplifies to (2/10) / (2/10 + 1/40) = (8/40) / (9/40) = 8/9.

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