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~