Consider a bag containing $3$ normal and $6$ doubly headed coins. A coin is selected randomly and tossed $3$ times and found to fall head-wise on all the $3$ occasions. Probability that the selected coin was a normal coin is $\dfrac {1}{17}$.

Using Bayes' theorem, the probability that the selected coin is normal given three heads is (1/9 * 1/8) / ((1/9 * 1/8) + (6/9 * 1)), which evaluates to (1/72) / (1/72 + 6/9) = 1/49. Because the calculated probability is 1/49 rather than 1/17, the given assertion is incorrect. Therefore, the assertion is incorrect but the reason is correct.

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