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A quiz has 3 multiple choice questions. For each question there are 4 choices and only one is correct. A certain student tries to guess all the answers. $$\\$$ Let Y be the random discrete variable that corresponds to the number of questions the student answers correctly.

What is the probability of the student getting: $$$$ a) All wrong $$\\$$ b) Failing one and getting the other two right $$\\$$ c) Getting two right and failing the other $$\\$$ d) All right

$$\\$$ $$\\$$

Normally I'use the probability (either $\frac{1}{4}$ or $\frac{3}{4}$ depending on the question) times the combinations, but I want to try something different.

I want to solve this problem this way:

  • find the total amount of sets for all the combinations
  • find the amount of sets required by the question
  • divide the latter by the first

I will exemplify with question a):

  • total combinations : $^4C_1*^4C_1*^4C_1 = 64$
  • amount of sets which the student gets all wrong = ???
  • divide: $$\frac{???}{64} = ???$$

My question is, how do I find the amount of specific sets out of the total 64 so I can solve the problem this way? Basically I want to find n and k in $^nC_k$, this being the amount of specific sets.

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  • $\begingroup$ For each of the three questions, there are three wrong answers and one correct one. To get all questions wrong, choose 1 of the 3 incorrect answers for each of the 3 questions. How many ways are there of doing this? $\endgroup$ – aduh Oct 16 '16 at 17:50
  • $\begingroup$ @aduh $^3C_1$ * 3? $\endgroup$ – Segmentation fault Oct 16 '16 at 17:56
  • $\begingroup$ @aduh actually, $^3C_1*^3C_1*^3C_1 = (^3C_1)^3$. Hey, this works! $\endgroup$ – Segmentation fault Oct 16 '16 at 18:04
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Let $(a,b,c) \in [1, 2, 3, 4]^3$ represent the answer of the student, with 4 being the right answer.

The triplets where the student gets all answers wrong are the elements of $[1, 2, 3]^3$, there are therefore $(C^3_1)^3$ possible triplets.

To number the triplet where the student gets only one answer right, you first choose which one is right ($C_1^3$ choices), you then have $C^1_1$ choice for the right answer and $C^3_1$ for the others, thus you have $(C^3_1)^3 C_1^1$ possible triplets.

And so on ...

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