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You are taking a multiple-choice test with n questions each of which has 4 alternatives. You have mastered 60% of the material. Assume this means that you have a 0.6 chance of knowing the answer to a random test question, and that if you don’t know the answer to a question then you randomly select among the four answer choices. Assume that this holds for each question, independent of the others, and assume that each correct answer gives 1 point and wrong answers give 0 points, the score is the sum of all points. For each answer define a random variable Xi (i=1,2,...,n) that takes the value 1 if the ith answer is correct and 0 otherwise. a.What is the probability that you answer a particular question correctly? b.What is your expected score on the exam? c.Write down a formula for the probability mass function (pmf) for one particular X, obtain the cumulative distribution function (CDF) for Xi and plot the CDF WORK: for my work so far I have A = Knowing the answer B = All choices are equal and C = Student answers correctly. P(A) = .6, P(B) = .25 I am looking for P(A|C)? = P(C|A)P(A)/P(C)? Other than that I am kind of lost

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  • $\begingroup$ for my work so far I have A = Knowing the answer B = All choices are equal and C = Student answers correctly. P(A) = .6, P(B) = .25 I am looking for P(A|C)? = P(C|A)P(A)/P(C)? Other than that I am kind of lost $\endgroup$ – White Mamba Feb 18 '18 at 23:22
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    $\begingroup$ You should include your attempt in the question itself. $\endgroup$ – N. F. Taussig Feb 18 '18 at 23:25
  • $\begingroup$ Ok sorry about that I am new to this site $\endgroup$ – White Mamba Feb 18 '18 at 23:26
  • $\begingroup$ What do you mean by "Write down a formula for the probability mass function (pmf) for one particular X"? $X$ is a random variable that can take on several different values. Do you mean $n$? $\endgroup$ – Remy Feb 18 '18 at 23:31
  • $\begingroup$ I believe for (a) you're looking for $P(C) = P(C | A)P(A) + P(C | B)P(B)$ (because $B = A^c$). $\endgroup$ – Pedro M. Feb 18 '18 at 23:31
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a) The probability that he gets a particular question correct is

$$p=\left(0.6\cdot1\right)+\left(0.4\cdot0.25\right)=0.7$$

b) This is a binomial distribution with mean $np$

c)

Assuming $X$ is a random variable taking on the value $1$ if the answer is correct and $0$ if the answer is incorrect...

$$ p_{X}(x)= \begin{cases} 0.7 & x =1 \\ 0.3 & x=0 \\ 0 & \text{otherwise} \end{cases} $$

d)

$$ F_{X}(x)= \begin{cases} 1 & x \geq 1 \\ 0.3 & 0\leq x\lt 1 \\ 0 & x \lt 0 \end{cases} $$

Note that even in the discrete case, we must account for all $x\in\mathbb{R}$

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  • $\begingroup$ for c don't you need 0 otherwise? Also for d don't you need x<0, 0<=x<1, and x>=1? $\endgroup$ – White Mamba Feb 18 '18 at 23:58
  • $\begingroup$ Yes, my mistake. I will fix that. $\endgroup$ – Remy Feb 19 '18 at 0:00
  • $\begingroup$ Also how do you get 1? I know that's the right answer im confused on how, for x>=1 $\endgroup$ – White Mamba Feb 19 '18 at 0:00
  • $\begingroup$ There is more than one correct way to notate the cdf. And what do you mean? $\endgroup$ – Remy Feb 19 '18 at 0:02
  • $\begingroup$ for the cdf you are saying that over one I thought it was .7 but why is .3 added in as well? Might be a dumb question but I dont know $\endgroup$ – White Mamba Feb 19 '18 at 0:06

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