Probability of multiple-choice answers in questions Assume a test in multiple-choice format taken by a student with no prior knowledge of the test subject and he is going to pick answers in a random way.


*

*What is the probability of him getting 100% in the test?

*What is the average probability of him getting atleast 40% in the test?

*Given N number of students take the test(assuming all pick answers at random), what is the average score?

*What effect do the variables number of questions, students and number of choices have in the overall average percentage of marks?

*Do the results considerably vary across tests?

*What would be a better strategy to get more marks - to pick answers at random or to select a single choice (such as A)


Assuming Q be the number of questions and Q, C be the number of choices and N the number of students and for this illustration let Q be 10, C be 3 and N be 1000, I arrived at the following
Question 1:
1/(C^Q). So 1/(3^10)
Question 2:
The probability of getting at least 40% is 1 minus probability of getting less than 4 questions correct. So 1-(1/(3^1+3^2+3^3))
Question 3:
1/C. I run a simulation program and got the results but can't mathematically prove or deduce it.
Question 4:
Deducing from 3 above, only C is going to have an impact on the average score irrespective of Q and N (assuming N to be greater)
Question 5:
Since only C is the major determinant of overall scores, the results aren't going to vary across tests
Question 6:
Randomly picking answers is better than picking a same choice. Again I ran a simulation and deduced it but cannot mathematically deduce it.
Am I right? How to mathematically deduce the answer for Question 3 and Question 6 
 A: Question 1
You're right, it is $1/C^Q$.
Question 3
By the law of large numbers, the average of the results obtained from a large number of trials should be close to the expected value.
The expected value, from the other side, is obviously $\frac{1}{C}$ (as, for each question, there is $\frac{1}{C}$ probability of getting $1$ and $\frac{C-1}{C}$ probability of getting $0$).
Question 4
Changing $C$ directly changes the expected average score. The number of questions and students affects the dispersion of the "average score".
Question 5
Results may vary considerably across tests, although more the number of questions and students is, the less likely considerable differences are.
For example, on the first test all students may accidentally guess all answers (which is possible, although unlikely); on the second test it is possible that none of the students will guess the right answer.
Question 6
As long as the correct answers are distributed uniformly, it doesn't matter which answer will you choose on a specific question, the probability of guessing is $\frac{1}{C}$. And, of course, the probability of getting the right answer for $n+1$-th question does not depend on what answer was chosen for $n$-th question. Both strategies are equivalent.
The results you got in your test run may be explained by e.g. non-uniform correct answers distributions (e.g. the second answer is correct for 50% of questions, while the first and the third answers are correct for 25% of questions) and checking against some unlucky answer (e.g. the first one). In such a case, choosing a random answer will give you 33.3% expected score, while always choosing the first answer will give you only 25% of expected score. 
A: Question 2: There is a problem with your calculation of the probability of getting at least $40$ percent. You decided, correctly, that it would be easier to find the probability of getting $\le 3$ questions right. Let $P(k)$ be the probability of getting $k$ questions right. The probability we get $\le 3$ is $P(0)+P(1)+P(2)+P(3)$. (You forgot about $P(0)$).  So the probability of getting $40$ percent or more is $1-[P(0)+P(1)+P(2)+P(3)]$.
It is easy to find $P(0)$. We must get the wrong answer $10$ times in a row. With $3$ choices, this probability is $(2/3)^{10}$.
What about $P(1)$? We need to get $1$ question right and the rest wrong. For any individual question, the probability you get it right is $(1/3)$, and the probability you get the rest wrong is $(2/3)^9$, for a probability of $(1/3)(2/3)^9$. However, which question you get right can be chosen in $\binom{10}{1}$ (that is, $10$) ways. The point is that you can get $1$ right and $9$ wrong in $10$ different ways. So the probability you get $1$ question right and the rest wrong is $\binom{10}{1}(1/3)(2/3)^9$.
What about $P(2)$? The probability you get two specific questions right and the rest wrong is $(1/3)^2(2/3)^8$.  However, the two questions you get right can be chosen in $\binom{10}{2}$ ways, so $P(2)=\binom{10}{2}(1/3)^2(2/3)^8$.
Similarly, $P(3)=\binom{10}{3}(1/3)^3(2/3)^7$. You need to learn about the Binomial Distribution, it will come up often in your work. I expect it has come up already, else this question would probably not be asked. If we repeat an experiment independently $n$ times, with probability of success each time equal to $p$, then the probability $P(k)$ of exactly $k$ successes is given by
$$P(k)=\binom{n}{k}p^k(1-p)^{n-k}.$$ 
Question 3: If $Q$ is the number of questions, and $C$ the number of choices, then the average score on a test is $\dfrac{Q}{C}$. The average score on any particular question is, as you surmised, $\dfrac{1}{C}$.
Question 4: The number of students does not affect the average score, but the number of questions on the test does, as does the number of choices per question. See the answer to Question 3.
Question 5: If Joe takes several such tests, all say $10$ questions, $3$ choices, Joe's score will vary considerably from test to test.  If we have a large class ($N$ big), then the average class score will not vary very much from test to test. 
Question 6: If the teacher, for each question, chooses at random where to place the right answer (A, B, or C), then it makes on average no difference whether we choose answers at random or pick A each time. If the teacher, as many teachers do, tries to have roughly speaking the same number of correct answers be A, B, or C, but given that constraint picks locations randomly, then again it doesn't make any difference to the average score. But if you pick all A, that will protect you against a very low score, and diminish the probability of a high score. A formal proof of these intuitively reasonable facts ("You can't beat Las Vegas) can be given, but is fairly difficult. 
