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Discrete R.V. Assume a student shows up for EGR 280 completely unprepared and the instructor gives a pop quiz. Assume there are 6 questions, every single question is multiple choice question. Assume that each question has five options (a, b, c, d, and e), the student selects the answer randomly. Also assume that whether one question is answered correctly is independent of whether or not another question is answered correctly.

a) For each single question, what it the probability (p) that the student answers correctly?
b) If the student correctly answers at least four questions, the student passes the quiz. What is the probability that the student passes the quiz?
c) What is the expected value (E(x)) for the quantity of questions that were answered correctly?

Answer :

Answer:

a) p=0.2

b) probability of passing is 0.01696

.

c) The expected value of correct questions is 1.2

Step-by-step explanation:

a) Since each question has 5 options, all of them equally likely, and only one correct answer, then the probability of having a correct answer is 1/5 = 0.2.

b) Let X be the number of correct answers. We will model this situation by considering X as a binomial random variable with a success probability of p=0.2 and having n=6 samples. We have the following for k=0,1,2,3,4,5,6

[tex] P(X=k) = \binom{n}{k}p^{k}(1-p)^{n-k} = \binom{6}{k}0.2^{k}(0.8)^{6-k}[/tex].

Recall that [tex]\binom{n}{k}= \frac{n!}{k!(n-k)!}[/tex] In this case, the student passes if X is at least four correct questions, then

[tex]P(X\geq 4) = P(X=4)+P(X=5)+P(X=6)=\binom{6}{4}0.2^{4}(0.8)^{6-4}+\binom{6}{5}0.2^{5}(0.8)^{6-5}+\binom{6}{6}0.2^{6}(0.8)^{6-6}= 0.01696 [/tex]

c)The expected value of a binomial random variable with parameters n and p is [tex]E[X] = np[/tex]. IN our case, n=6 and p =0.2. Then the expected value of correct answers is [tex]6\cdot 0.2 = 1.2[/tex]

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