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.
Multiple choice questions (MCQs) are common in examinations over here in Singapore. A set of, say, $40$ questions are given to students, and each is accompanied with a list of $4$ choices of answer, $A, B, C$ or $D$.
Suppose a student did not study for a particular test. Out of desperation, he decides to "guess" a choice for each question to secure at least some points. He considers two possible strategies:
- For each question, he picks a random choice as his answer.
- He picks a random choice, say, $B$ and uses it for all his answer.
Assuming the correct answers are randomly distributed and his guess is completely random, which strategy would give a higher probability of securing more correct answers than the other?
The variable here is number of correct answers, i.e. let it be $X$.
1) $$\rm E[X]=\sum_{k=0}^{40} P(X=k)k=\sum_{k=0}^{40}\binom{40}k\left(\frac14\right)^k\left(\frac34\right)^{40-k}k=10$$
This is $25\%$. Now the standard deviation is:
$$\rm \sigma[X]=\sqrt{\sum_{k=0}^{40}(k-10)^2\binom{40}k\left(\frac14\right)^k\left(\frac34\right)^{40-k}}=\sqrt{\frac{15}2}\approx2.73861$$
2) $$\rm E[X]=\sum_{k=0}^{40} P(X=k)k=\sum_{k=0}^{40}\binom{40}k\left(\frac14\right)^k\left(\frac34\right)^{40-k}k=10$$
This is $25\%$. Now the standard deviation is:
$$\rm \sigma[X]=\sqrt{\sum_{k=0}^{40}(k-10)^2\binom{40}k\left(\frac14\right)^k\left(\frac34\right)^{40-k}}=\sqrt{\frac{15}2}\approx2.73861$$
Sorry, but do you know why I did that? Because $\rm P(X=k)$ is constant for both because of inter-independence of questions.I would further say that any strategy you device would all get you same results. I can't prove it, but you can call it my intution. I think like publishing this theory of independent results! :D
Best Answer
The probability of getting exactly $k$ successes in $n$ trials is given by $$ f\left(k;n,p\right)=\left(\begin{array}{c} n\\ k \end{array}\right)p^{k}\left(1-p\right)^{n-k}. $$ Here, $k=8$, $n=10$ and $p=1/4$. Plug in and crunch. If however, you want to know the probability that Jim gets at least $8$ correct, you simply need to sum $$ f\left(8;10,1/4\right)+f\left(9;10,1/4\right)+f\left(10;10,1/4\right). $$