The multiplication Principle exercise I’m on the quest to learning Statistics and Probability as part of my masters degree and one of the exercises proposed by the lecturer is giving me some trouble to solve.
The problem has three parts and I’ve already solved the first two parts but the final question I cannot get to a way of solving it and would like to know if anyone can help. The problem is as follows:
“A history test contains 6 questions and lists 3 possible answers ( only 1 is correct in each case) for each question. A student who has done no work in the subject selects his answers by guesswork. In how many ways may the answers be selected? Do you think this student had much chance of scoring 6 correct? Of scoring 3 correct?”
For the first one I know it is 3^6 = 729
For the second I know it is 1^6 = 1
However, the nominated result for the 3 good out of the 6 is 540 but I cannot get a reasonable calculation for this. The method we are using is the box method but am not sure how to put it down on paper and not sure whether that 540 answer is correct.
Any light on this solve shall be appreciated.
Christian
 A: For the probability of all 6 correct, there is $1/729.$
You are mixing probabilities and numbers of outcomes, but
it seems you're on the right track for that one.
For the probability of getting exactly 3 out of 6 correct,
here is one argument. One way to do that is to get the first three correct and the last three incorrect.
Multiplying probabilities of independent events, you'd have $$ (1/3)(1/3)(1/3)(2/3)(2/3)(2/3) = 8/3^6 = 8/729.$$
However, there are 20 possible ways to arrange the correct
and incorrect answers: CCCIII, CCICII, ..., IIICCC.
The number of arrangements is ${6 \choose 3} = \frac{6!}{3!\cdot 3!} = 20.$
So, the total probability of getting some three correct out of six questions is $20\frac{8}{3^6} = \frac{160}{729}.$
I suppose you will learn about the binomial distribution in your course soon. It gives the probabilities of 
getting exactly $k = 0, 1, 2, \dots, 6$ correct answers out of 6 by random guessing.
The general formula for getting exactly $k$ correct is
$$ {6 \choose k}(1/3)^k(2/3)^{6-k}.$$
Your question deals with the cases $k = 6$ and $k = 3.$
Here is a table of the probabilities of all possible values of $k,$  rounded to 5 decimal places.
 k     prob
 0  0.08779
 1  0.26337
 2  0.32922
 3  0.21948   # 160/720 
 4  0.08230
 5  0.01646
 6  0.00137   #   1/729

If you add all seven probabilities together, you get 1 (except for rounding error), so we have accounted for all the probability.
