How to find out the probability? The question is:
A true or false test contains eight questions. If a student guesses the answer for each question, find the probability:
(a) 8 answers are correct. (b) 7 answers are correct. (c) 6 answers are correct. (d) at least 6 answers are correct.
I proceed as that total number of questions are eight, each question may be true or false, so $n(S)=16$, I don;t know whether am I correct to find the sample space or incorrect?
Also I don't know how to proceed and calculate the other parts of the questions?
You have 8 questions, each a choice from two answers. &nbsp; The state space contains $2^8$ distinct outcomes selected without bias (ie having an uniform probability distribution).

Thus the probability for each individual outcome to occur is $1/256$.


  (a) 8 answers are correct.


Obviously only one outcome corresponds to this event. 


  (a) 7 answers are correct.


We must count the ways to select seven from eight questions to give true answers. 

So there are $\tbinom 87$ or simply $8$ outcomes in this event. 

Thus the probability is $8/256$, or simply $1/32$.


  …


You can do the rest.
The answer for that question is given by the formula $$\Big(\frac{1}{n}\Big)^k$$ then multiply it to the number of combinations.

I will explain why. For the first question there will be 2 possiblities - a True or a False(let's just use T and F). For the 2nd question there will be 2 more for each possible event of the first. TT, TF, FT,FF. For the third there will be 2 more for each previously mentioned samples.

a. 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 = 256 or $(\frac{1}{2})^8 = \frac{1}{256} * \frac{8!}{8!(8-8)!}$ = $\frac{1}{256}$

b. 2 x 2 x 2 x 2 x 2 x 2 x 2 = 128 or $(\frac{1}{2})^7 = \frac{1}{256}  * \frac{8!}{7!(8-7)!}$ = $\frac{1}{32}$

c. 2 x 2 x 2 x 2 x 2 x 2 = 128 or $(\frac{1}{2})^6 = \frac{1}{256} * \frac{8!}{6!(8-6)!}$ = $\frac{7}{64}$