If there are $n$ questions on the test, then the probability of answering exactly $k$ correctly, if answers are chosen at random, is
$$\binom{n}{k}\left(\frac{1}{3}\right)^k \left(\frac{2}{3}\right)^{n-k}.$$
And it is the Muliplication Rule. To make typing simpler, I will take $k=2$, $n=5$. Write $C$ for correct, $N$ for not correct.
By the Multiplication Rule, the probability of $CCNNN$ (first two right, next three wrong) is $(1/3)^2(2/3)^3$. But we can also get two right, three wrong in several other ways, like $CNCNN$, $CNNCN$, and so on. Each has probability $(1/3)^2(2/3)^3$.
How many such strings are there? We have to choose $2$ places from the $5$ available to put a $C$ into. There are $\binom{5}{2}$ ways to do this, giving total probability $\binom{5}{2}(1/3)^2(2/3)^3$.
The same reasoning gives the general formula quoted above.
Remark: If there are only $5$ questions on the test, and we want the probability of answering all $5$ correctly, the probability is $(1/3)^5$, which is nowhere near the number quoted.
Because the formula for the probability has only $3$'s in the denominator, any probability we compute will have to be of shape $\frac{m}{3^e}$, where $m$ and $e$ are non-negative integers. The number $\frac{45}{118}$ is not of that shape, so cannot be the answer for the problem, or for any closely related problem.
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 simplest way to do (a) is to find the probability of answering 0 or only 1 correctly, then subtracting that from 1. With four possible answer to each problem, the probability of answering one correctly by choosing at random is 1/4 and the probability of answering incorrectly is 3/4. So the probability of answering all 8 questions incorrectly (0 correctly) is $(3/4)^8= 0.31640625$. The probability of answering a specific question correctly and the other seven incorrectly is $(1/4)(3/4)^7= 0.033370971679685$. The probability any one of the seven is correct and the other 3 incorrect is that times 8, 0.2669677734375. The probability that either all 8 are incorrect or one is correct and the other 7 are incorrect is 0.31640625+ 0.2669677734375= 0.5833740234375. The probability of answer 2 or more correctly is 1- 0.5833740234375= 0.4166259765625.