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?

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    $\begingroup$ Do you know about the binomial distribution? $\endgroup$ – saulspatz Apr 1 '20 at 14:39
  • $\begingroup$ Your state space can be seen as a vector with 8 entries being either 0 or 1 to represent false or true. So you are looking for all possibilities $S = \{(r_1,r_2,...,r_8)| r_i\in\{0,1\})\}$, which corresponds to a size of $n(S) = 8^2 = 64$ (8 entries, 2 possibilities each) $\endgroup$ – Jfischer Apr 1 '20 at 14:41
  • $\begingroup$ You can read up on Bernoulli trials and binomial distribution as stated by @saulspatz $\endgroup$ – Adnan Selimovic Apr 1 '20 at 14:42
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    $\begingroup$ @Jfischer You mean $2^8=256$ possibilities, don't you? $\endgroup$ – saulspatz Apr 1 '20 at 14:46
  • $\begingroup$ @Jfischer How $2^8=256$ please explain brother! $\endgroup$ – Noor Aslam Apr 1 '20 at 14:58

You have 8 questions, each a choice from two answers.   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.


Let us look at this from beginning. As you must know, P(A)= (no. of favorable outcomes)/(no. of total outcomes). Now we can solve this question with this basic formula. As there are two equally likely cases for each of the 8 questions that is, it would be true or false, we now have a total of 2⁸=256 outcomes.

(a) All the question can be correct in only one way, thus P(8 correct)=1/256 (b) 7 question can be correct in (⁸₇)=8 ways, hence P(7 correct)=1/32 (c) 6 question can be correct in (⁸₆)=28 ways, hence P(6 correct)=7/64 (d) atleast 6 questions can be correct in (⁸₆)+(⁸₇)+(⁸₈)=37 ways, hence P(atleast 6 correct)=37/256


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 $(\frac{1}{2})^6 = \frac{1}{256} * \frac{8!}{6!(8-6)!}$ = $\frac{7}{64}$

  • $\begingroup$ Completely wrong. Look up binomial distribution to see why. $\endgroup$ – David G. Stork Apr 2 '20 at 0:16
  • $\begingroup$ You are calculating the probability for getting the first $k$ of the questions correct. $\endgroup$ – Graham Kemp Apr 2 '20 at 0:18
  • $\begingroup$ Damn I just realized I forgot to count the combinations sorry. $\endgroup$ – ライガン アンドレー イマニューエル Apr 2 '20 at 0:19

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