A box contains fair coins and unfair coins. When one of the unfair coi

The probability of landing 1H1T for the fair coin is 50%*50% + 50%*50% = 50%.
The probability of landing 1H1T for the unfair coin is \(\frac{2}{3}*\frac{1}{3} + \frac{1}{3} * \frac{2}{3 }= \frac{4}{9}\). For now let us focus on the numbers 50% and 4/9.

If we had one fair coin and one unfair coin, then we would have a \(\frac{(50\%) }{ (50\% + 4/9)} \) chance that the coin is fair based on the 1H1T result. This value is a bit more than 50% which makes sense, if the random toss result is 1H1T then it is more likely we got the fair coin but only a tiny bit more since we still have a good chance of the unfair coin rolling a 1H1T. If the result was 2H it is more likely we get the unfair coin.

However we have the probability mentioned above at 50% instead, this means there are relatively more unfair coins and less fair coins. Let us use x for fair coins and y for unfair coins. Keeping the weights 50% and 4/9 but scaled with x fair coins and y unfair coins, we would get:

\(\frac{50\% * x }{ (50\%*x + (4/9) * y)} = 50\%\).

After a bit of arrangement, we would get: \(50\%*x = \frac{4}{9}y\).

And \(4.5*x = 4y\), or \(9x = 8y\). Finally the smallest solution here is y = 9 and x = 8, thus a total of 17 coins.

Ans: C

Note: For more info see Bayes' Theorem, this question is clearly out of GMAT scope but it can be interesting for probability enthusiasts.