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13 Sep 2020, 03:29
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C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
85%
(hard)
Question Stats:
47% (02:32) correct
53%
(02:21)
wrong
based on 83
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A box contains fair coins and unfair coins. When one of the unfair coins is flipped, the probability that it lands heads is \(\frac{2}{3}\). A coin is randomly drawn from the box and flipped twice. One of these flips lands heads, and the other one lands tails. The probability that the coin was fair is exactly \(\frac{1}{2}\). Find the minimum number of coins in the box.
A. 13
B. 15
C. 17
D. 19
E. 21
A. 13
B. 15
C. 17
D. 19
E. 21
The probability that the coin was fair is exactly 1/2
I think if the probability of getting fair and unfair coins is same 1/2 then total no of coins must be even 2*(same no of fair and unfair coins)
I suppose this question is wrong please correct me if I am wrong.
I think if the probability of getting fair and unfair coins is same 1/2 then total no of coins must be even 2*(same no of fair and unfair coins)
I suppose this question is wrong please correct me if I am wrong.
13 Sep 2020, 15:47
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nick1816
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.
