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Question: A multiple choice exam has 60 questions. Each question has 4 possible answers and only 1 answer out of the 4 possible answers is correct. To receive an A grade, one must answer 95% and above of the questions correctly. We know that 54 questions were answered correctly. What is the probability of receiving an A grade (rounding o to 3 decimal places), if one were to guess the remaining questions?

(Note: Binomial theorem/distribution not to be used)

For the above question, I have worked out that in order to get an A grade, we'd need at least $\frac{95}{100} \times 60 = 57$ questions correct. Thus that means we have to get at least 3 out of the remaining 6 questions correct.

This leaves me with:
1) Correct Correct Correct Wrong Wrong Wrong
2) Correct Correct Correct Correct Wrong Wrong
3) Correct Correct Correct Correct Correct Wrong
4) Correct Correct Correct Correct Correct Correct

I know that for case (1) that the probability for any particular combination is $(\frac{1}{4})^3(\frac{3}{4})^3$ but the order does matter in the sense that getting the last 3 correct versus getting the first 3 correct are separate events, is that right?

How should I go about solving this question with the information I have analysed thus far with just conditional probability and no binomial theorem involved?

Thank you~

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  • $\begingroup$ "the order does matter in the sense that getting the last 3 correct versus getting the first 3 correct are separate events, is that right?" Yes, it is. For each possible order, the probability is the same, and there are $\binom63$ possible orders. Therefore the contribution from 1) is $\binom63(\frac14)^3(\frac34)^3$. $\endgroup$ – Arthur Nov 16 '16 at 9:50
  • $\begingroup$ @Arthur Thank you for the confirmation! I believe I understand the approach to the question now from both your reply and Robert's. $\endgroup$ – stephchia Nov 16 '16 at 10:17
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Hint. We need at least 3 correct answers out of $6$ (and the order does matter). Hence the probability is $$p=\sum_{k=3}^6\binom{6}{k}(\frac{1}{4})^k(\frac{3}{4})^{6-k}=\frac{20\cdot 3^3+15\cdot 3^2+6\cdot 3+1}{4^6}=\frac{347}{2048},$$ where $k$ is the number of correct answers, $\binom{6}{k}$ is the number of ways to select them.

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  • $\begingroup$ Well, there's a note saying "Binomial theorem/distribution not to be used". Not really sure what's the purpose of this restriction though... $\endgroup$ – barak manos Nov 16 '16 at 9:48
  • $\begingroup$ @barak manos You are right. This restriction is quite bizarre. $\endgroup$ – Robert Z Nov 16 '16 at 10:06
  • $\begingroup$ @RobertZ Thank you for the speedy reply, I believe I understand the reasoning behind your working now! $\endgroup$ – stephchia Nov 16 '16 at 10:14
  • $\begingroup$ @barakmanos The reason was that, I believe that this can be done by finding (1 - P(x smaller or equals 2)) setting a binomial of Bi(6, 0.25). But I did not want to resort to that. But thank you! $\endgroup$ – stephchia Nov 16 '16 at 10:16

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