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So I know this may seem like a simple question, but i'm a student and the difference seems trivial but confusing to me.

So basically you have 2 types of students in a class, guessers and swaters. They all sit an exam with n number of questions, swaters have 100% chance of getting the answers right and guessers have 50% chance as all questions are yes/no. So the lecturer sets the number of questions n to 7 to reduce the chance of a guesser getting 100% in the test to less than 1%.

Now the questions is, since all swatters will get 100% no matter what n is, the guessers can be identified as any student who doesn't get 100%. So if there are 20 guessers in the class what is the probability of catching all the guessers.

So in my mind the simple way to do this is the binomial random variable $X\sim B(7,0.5)$ this is for $P(X=7)=(0.5)^7$ so probability of catching the guesser is 1-(0.5)^7. So then the probability of catching all the guessers is $(1-0.5^7)^{10}$.

But what if you define a random variable for the total number of questions answered correctly by all guessers $Xtot \sim B(140,0.5)$. Then catching them all probability is $(1-0.5^{140})$. So which is correct, because you get different answers. And why is it correct? I have to go with the first way, but I am unsure as to the reason.

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  • $\begingroup$ Actually I think I get it, the second one is actually calculating the probability that not all of them are caught, not that all of them are caught. Is this correct? $\endgroup$ Commented May 6, 2015 at 8:42
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    $\begingroup$ Do you mean swotters? $\endgroup$
    – Glen_b
    Commented May 6, 2015 at 9:11

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$1-.5^{140}$ is the probability of catching any guessers, i.e. The probability that any guesser gets a question wrong.

Thinking only about the total number of questions correct by the guessers cannot tell you how many guessers have been caught in general, because it doesn't distinguish between different people: for example, that model cannot tell the difference between three guessers each getting 7 questions wrong and the others all correct (21 wrong questions; three guessers are caught), versus each guesser getting one question wrong except for one that gets two (21 total, and all 20 guessers are caught).

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