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Bob took a quiz consisting of 500 questions with two options: Yes and No. Bob did not prepare for the quiz. Flustered, he hastily reached for a fair coin in his wallet and started to toss this coin for answers. Whenever the coin landed heads, he shaded Yes; whenever the coin landed tails, he shaded No. It turned out that 40% of the questions have Yes as the correct answer, and the rest have No as the correct answer. Among those questions that Bob shaded Yes, roughly how many percent did he get correct?

A) 30% B) 40% C) 50% D)60% E) 70%

My thought is that of the questions he put as Yes, the chance of getting the right answers is equal to 40%. But this is just a guess and I'm not sure of the explanation or if this is the right answer.

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    $\begingroup$ Makes no difference what the true percentage is. Bob has a $\frac 12$ chance on each, so he should get about $50\%$ right. $\endgroup$ – lulu Feb 15 '17 at 14:14
  • $\begingroup$ Each time a yes question appears there is a 50% chance he will answer yes. The other questions that appear we don't need to consider because we won't count them. $\endgroup$ – mathreadler Feb 15 '17 at 14:27
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    $\begingroup$ Note: I misread the question. I thought you were asking what Bob's expected score is. Any given question has a $.4$ chance of being correctly answered Yes. Therefore Bob should get $40\%$ correct whenever he guesses Yes. Similarly, he gets $60\%$ correct when he guesses No. Therefore, just to be complete, he get $.4\times 12+.6\times 12=.5$ right on average. $\endgroup$ – lulu Feb 15 '17 at 14:34
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    $\begingroup$ @mathreadler What you want to say is: Each time he answers yes, there is a 40% chance he is correct. The other questions that appear we don't need to consider because we won't count them. $\endgroup$ – Matjaž Krnc Feb 15 '17 at 14:34
  • $\begingroup$ @KevinWells As you'll see from a subsequence comment of mine, I was misreading the question (as did many other responders). $\endgroup$ – lulu Feb 15 '17 at 18:13
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It's 40%. Here's an equivalent statement:

Take a random subset $S$ among all the questions (can also be called sampling) - each question is taken with probability $\frac{1}{2}$. Since questions are chosen uniformly at random it still holds that 40% of the questions have Yes as the correct answer.

But $S$ corresponds to the $\mathtt{Yes}$-subset (from your question).

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    $\begingroup$ @LaarsHelenius In fact this is right. The question is subtly worded and doesn't mean what almost everyone thought at first. $\endgroup$ – Ethan Bolker Feb 15 '17 at 14:41
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    $\begingroup$ @EthanBolker The question asks "Among those questions that Bob shaded Yes, roughly how many percent did he get correct?"; what is subtle about that wording? $\endgroup$ – ShreevatsaR Feb 15 '17 at 21:04
  • $\begingroup$ @ShreevatsaR OK maybe not subtly worded, but compactly worded and easy to misread. $\endgroup$ – Ethan Bolker Feb 15 '17 at 21:43
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I think Bayes' Theorem makes this most clear. The probability of Bob answering Yes on any question is $P(Y)=0.5$, the probability of Bob being correct on any question is $P(C)=0.5$, and the probability that Bob answers Yes given that he answered correctly is the proportion of questions with Yes as the answer, $P(Y|C)=0.4$. This leads to:

$$P(C|Y)=\frac{P(Y) \cdot P(Y|C)}{P(C)}=\frac{0.5 \cdot 0.4}{0.5}=0.4$$

Thus, the estimated percentage of correctly answered questions that Bob shaded Yes is $P(C|Y)=0.4=40\%$

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The amount he will get correct is independent of what the distribution of right and wrong answers is.

You can imagine this by assuming that all the correct answers are "No" ($0\text{%}$ Yes, $100\text{%}$ No). By the reasoning you've made, all the answers obtained from the coin will be wrong, which is obviously incorrect.

Therefore, the probability should be $50\text{%}$ since it is a fair coin.


Edit: As @lulu has pointed out, the above is only applicable if both the "Yes" and "No" answers are considered.

However, since you've mentioned: "Among those questions that Bob shaded yes", the answer should be $40\text{%}$ as other answers are pointing out.

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    $\begingroup$ Note: Like you, I misread the question. The OP is only asking about the times Bob guesses Yes. That narrower question does depend on the true distribution. To take an extreme case, suppose every single question is correctly answered Yes. Then Bob gets $100\%$ of his Yes guesses right. $\endgroup$ – lulu Feb 15 '17 at 14:35
  • $\begingroup$ Oh, yes thank you for pointing this out. Then it should be $40\text{%}$. $\endgroup$ – projectilemotion Feb 15 '17 at 14:37
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    $\begingroup$ Please don't delete this incorrect answer. Instead edit it to make your misreading clear. That way others can learn from it. I'll upvote if you do that. $\endgroup$ – Ethan Bolker Feb 15 '17 at 14:39
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    $\begingroup$ I agree with @EthanBolker . The misreading is extremely seductive (I made the exact same error) so I think we can expect many people to fall into the same trap. Very much worth pointing out. $\endgroup$ – lulu Feb 15 '17 at 14:41
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    $\begingroup$ @mathreadler The question is stated perfectly clearly: "Among those questions that Bob shaded Yes, roughly how many percent did he get correct?" What is unclear about it? I believe only someone who didn't read that sentence would misunderstand the question. $\endgroup$ – ShreevatsaR Feb 15 '17 at 16:33

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