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27 Jul 2026, 00:57
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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)!
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23% (01:09) correct
77%
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Eirini has two fair coins and one coin with heads on both sides. She chooses one of the three coins at random and tosses it. If the coin comes up heads, what is the probability that the opposite side is tails?
(A) \(\frac{1}{3}\)
(B) \(\frac{2}{5}\)
(C) \(\frac{1}{2}\)
(D) \(\frac{5}{8}\)
(E) \(\frac{2}{3}\)
(A) \(\frac{1}{3}\)
(B) \(\frac{2}{5}\)
(C) \(\frac{1}{2}\)
(D) \(\frac{5}{8}\)
(E) \(\frac{2}{3}\)
27 Jul 2026, 05:30
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This is the classic "two-headed coin" Bayes trap dressed up a little differently, and I love that literally 0% of people have gotten it right so far on this thread.
1. Set up the coins. Two fair coins (each has one H, one T) and one double-header (both sides H). Eirini grabs one at random and flips.
2. The instinct almost everyone has is "well 2 of the 3 coins are fair, so the answer's obviously something with a 2 or 3 in it," and picks A or B. That's the trap. You're being asked a conditional probability question (given heads showed up), not a plain "what fraction of coins are fair" question.
3. Easiest way I know to do this without the formal Bayes formula: think of it as 6 total faces instead of 3 coins. Fair coin 1 = H,T. Fair coin 2 = H,T. Double coin = H,H. Every face is equally likely to end up on top, 1/6 each.
4. Out of those 6 faces, 4 are heads. Of those 4 heads-faces, only 2 have a tails on the flip side (the two fair-coin heads). The double-header's heads never has a tails on the back.
5. So P(opposite is tails given heads showed) = 2 out of 4 = 1/2. Answer C.
The reason this trips people up is that seeing heads is actual evidence, it should make you suspect the double-header a bit more than its plain 1/3 odds, and that pulls your answer down from 2/3 toward 1/2.
Takeaway: whenever a probability question gives you a result and then asks about what caused it, that's your cue you're doing Bayes whether the problem says so or not.
1. Set up the coins. Two fair coins (each has one H, one T) and one double-header (both sides H). Eirini grabs one at random and flips.
2. The instinct almost everyone has is "well 2 of the 3 coins are fair, so the answer's obviously something with a 2 or 3 in it," and picks A or B. That's the trap. You're being asked a conditional probability question (given heads showed up), not a plain "what fraction of coins are fair" question.
3. Easiest way I know to do this without the formal Bayes formula: think of it as 6 total faces instead of 3 coins. Fair coin 1 = H,T. Fair coin 2 = H,T. Double coin = H,H. Every face is equally likely to end up on top, 1/6 each.
4. Out of those 6 faces, 4 are heads. Of those 4 heads-faces, only 2 have a tails on the flip side (the two fair-coin heads). The double-header's heads never has a tails on the back.
5. So P(opposite is tails given heads showed) = 2 out of 4 = 1/2. Answer C.
The reason this trips people up is that seeing heads is actual evidence, it should make you suspect the double-header a bit more than its plain 1/3 odds, and that pulls your answer down from 2/3 toward 1/2.
Takeaway: whenever a probability question gives you a result and then asks about what caused it, that's your cue you're doing Bayes whether the problem says so or not.
28 Jul 2026, 04:09
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Edskore
Though I was unaware of application of Bayes formula in this question, I used the same approach, but got different answer.
I solved it like this:
Total possibilities = 6 (since there are 4 heads and 2 tails)
Given that one heads showed up, probability that flip side has another "heads" = 1/3
So, probability that flip side has tails = 1 - 1/3 = 2/3
What am I missing here?
