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In a $12$-item multiple choice examination (each with four choices of which only one is correct), what is the probability of getting only $4$ mistakes assuming that no question is left unanswered?

I used binomial distribution since each of the four choices in each item is equally likely to be picked on and the choices made on each question are independent.

I got $P(X=4)$ $=$ $0.002389848232$

I would like to verify if my method and answer is correct.

David G. Stork
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PRD
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  • But the answer in yiur solutiin si different from mine. – PRD Oct 13 '19 at 07:15
  • This is my solution, since the probability of my success is having a mistake. $${12 \choose 4} \left(\frac{3}{4}\right)^4 \left(\frac{1}{4}\right)^8$$ – PRD Oct 13 '19 at 07:20

2 Answers2

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Binomial probability gives:

$${12 \choose 4} \left(\frac{3}{4}\right)^4 \left(\frac{1}{4}\right)^8$$

which matches your answer.

Toby Mak
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  • This is my solution, since the probability of my success is having a mistake. $${12 \choose 4} \left(\frac{3}{4}\right)^4 \left(\frac{1}{4}\right)^8$$ – PRD Oct 13 '19 at 07:18
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    You are correct. I typed my answer wrong, which must explain why I got it right the first time :) – Toby Mak Oct 13 '19 at 07:20
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    Why this one will not work? ${12 \choose 4}$ $÷$ $4^1^2$? – PRD Oct 13 '19 at 07:22
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    Because this only accounts for the incorrect answers. To ensure that the other answers are all correct, you have to multiply by some factor (still working it out). I get that this factor should be $80$ by Wolfram Alpha. – Toby Mak Oct 13 '19 at 07:24
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    @MRA don't worry about the combinatorial thing I mentioned; I left something out (still trying to find what it was) that made me incorrect. Don't worry, you are fine. That number is all of the possible choices of 4 questions out of 12. Going to continue to try to figure out what else you need to multiply by to get the actual number. – Shinaolord Oct 13 '19 at 07:27
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    The correct answer should be $\frac{3^4 \cdot {12 \choose 8}}{4^{12}}$. There are $4$ incorrect answers with $3$ possible options each, and for the remaining $8$ questions, although there is only $1$ possible answer (the correct answer), there are ${12 \choose 8}$ ways to arrange the $8$ questions. – Toby Mak Oct 13 '19 at 07:33
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    @Toby Mak. Actually that was my answer in my first solution. After further review of the problem, I figured out that it is a Binomial Experiment. I am also thinking of what concept I missed in the first solution. – PRD Oct 13 '19 at 07:33
  • @TobyMak and the 3^4 comes from? Since there are 4 different answer per questions, and 3 are the incorrect, we multiply by it? – Shinaolord Oct 13 '19 at 07:35
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    Yes, that's correct. – Toby Mak Oct 13 '19 at 07:35
  • @MRA Your solution was correct. This is just a different way to arrive at the same answer. – Toby Mak Oct 13 '19 at 07:36
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    I will admit I replied too hastily, trying to show an alternative route. So sorry for confusing you @MRA – Shinaolord Oct 13 '19 at 07:36
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    Actually $\frac{{12 \choose 4} \cdot 3^4}{4^{12}}$ is correct as well since ${12 \choose 4} = {12 \choose 8}$. Once you choose the $4$ incorrect answers, the remaining answers must all be correct, and then you have the $3^4$ possible ways to choose the incorrect answers. – Toby Mak Oct 13 '19 at 07:38
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    Yes, there are only $12 \choose 4$ ways to choose the correct answers. – Toby Mak Oct 13 '19 at 07:49
  • Let us continue this discussion in chat. – Toby Mak Oct 13 '19 at 07:50
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Thanks for the assistance of @TobyMak for helping me (us) figure this out

There is a combinatorial approach to this problem. First, the total number of possible sets of solutions is $4^{12}$. Of these, the number of sets of 4 different problems is $\binom{12}{4}=\binom{12}{8}$.

Then, we turn to identifying the number of those sets of 4 questions that are all incorrect. For each question we choose, 3 of the possible answers to the question are incorrect, while only one is correct. Since we are choosing 4 questions, we multiply by 3, four times, to get the number of possible incorrect sets of 4 questions. this leads us to the following:

$$N(\text{4 Incorrect Questions out of 12})= 3^4\binom{12}{4}$$ and accordingly:

$$N(\text{4 Correct Questions out of 12})= \binom{12}{4}$$ This is the same as the number of options of 4 questions out of the 12.

(Side note: This actually tells us the total number of possible combination of either 4 correct or 4 incorrect answers, of 4 out of 12 questions is $(1+3^4)\binom{12}{4}$)

Now then, the probability will be $$100\frac{N(\text{4 incorrect questions out of 12})}{4^{12}}=100*3^4 \frac{\binom{12}{4}}{4^{12}}=\frac{1,002,375}{4,194,304}\approx 0.238985 \%$$, because we have $4^{12}$ possible choices and $N$ possible combinations of 4 incorrect answers.This is equal to the answer you gave via the binomial distribution.

Shinaolord
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