probabilities on multiple choice questions I am setting a series of multiple choice questions. I have a range of option answers from 2 to 7. Some questions only have a single correct answer and others have more. 
I can tell the person taking the test what number are correct, or I can just tell them to 'select all that are correct'.
Is there a difference in the probability of someone guessing correctly between the two scenarios?
I am, by the way, absolutely clueless at maths so please dumb your answer down to the lowest common denominator! eg Yes- if you tell them there are two correct answers it is xx:yy and if you don't tell them it is aa:bb.
Thank you.
 A: Suppose you tell them just two are correct and they guess two at random. There are $10$ ways to choose two from five (the number is small enough so you can just count them - no fancy math) and just one of the ten is right, so the probability of a correct guess is $1/10$.
If you want to count getting one or two of the two right the calculation is harder; I don't think that's what you're asking.
If you don't tell them two are right then you have to make an assumption about their guessing strategy. If they think there might be anywhere from one to five correct answers they will have $2^5 -1 = 31$ possible guesses one of which will be right, for a probability of $1/31$.
If I were taking the test I might think two or three right was the answer and would limit my guesses appropriately.
In any case giving the test taker the information will increase the probability that a guess will be correct.
A: Let's say the question's options are ${A, B, C, D, E}$. 
If the person was told there were only two options, he would guess from:
$\{A,B\},\{A,C\},\{A,D\},\{A,E\},\{B,C\},\{B,D\},\{B,E\},\{C,D\},\{C,E\},\{D,E\}$
As there is 10 options, and one of them is correct, his chance of guessing correctly is $\frac{1}{10}$
Note that we do not count rearrangements such as $\{B,A\}$ because it does not matter what order the person selections the options in. 
Now, if he had to randomly guess, he would have to pick from all the subsets of $\{A, B, C, D, E\}$ - essentially the power set of that set. 
Recall that the size (cardinality) of a power set is equal to $2^n$, where $n$ is the size of the original set. In this case, $n=5$, so there are $2^5=32$ options to pick from. Thus, his chance of guessing correctly is $\frac{1}{32}$.
Note - this includes the fact that the person might think none of the answers are correct. If at least one option is correct, then the chance is $\frac{1}{31}$
A: Another way of expressing the answer, assuming that the student knows that there are two correct answers out of five, is
The probability of getting both correct by pure guess is 
$$\frac{2}{5} \times \frac{1}{4}=\frac{1}{10}$$
(2 in 5 chance of picking the right answer on the first pick. On the second pick the student has a 1 in 4 chance of picking the right answer because one of the correct answers is gone.)
The probability of getting just the first answer correct by pure guess is 
$$\frac{2}{5} \times \frac{3}{4}=\frac{3}{10}$$
(On the second pick the student has a 3 in 4 chance of picking the wrong answer because one of the correct answers is gone.)
The probability of getting just the second answer correct by pure guess is 
$$\frac{3}{5} \times \frac{2}{4}=\frac{3}{10}$$
(3 in 5 chance of picking the wrong answer on the first pick. On the second pick the student has a 50:50 chance of picking the right answer because neither of the correct answers is gone.)
The probability of getting no answers correct by pure guess is 
$$\frac{3}{5} \times \frac{2}{4}=\frac{3}{10}$$
(On the second pick the student has a 50:50 chance of picking the wrong answer because neither of the correct answers is gone.)
Perhaps the student should get more than half the marks for getting both answers correct. 
