Question

Problem 7. A test contains n = 3 true or false questions. Any given student answers each question correctly with a certain probability called the "success rate"), independently acorss questions. In the class, 80% of the students know the materials quite well and have a success rate of p1 = 0.9, so the number of correct answers they get follows Bn, Pı). The rest of the students do not know the material at all, and they will do a coin toss to answer each the questions with a success rate of P2 = 0.5 on each, So the number of correct answers for these students follows Bn, p2). For a randomly chosen student, let X denote which group the student belongs to, and Y denote the number of correctly answered questions. Then X and Y can be modeled using the following hierachical framework: X=21 1, 12, if the student knows the material well, with probability 80% if the student does not know the material at all, with probability 20%, and given X = 1 or 2, the conditional distribution of Y is Y|X = I ~ B(n, pz), where pz = P1 = 0.9 if x = 1 and pe = P2 = 0.5 if x = 2. 1. Suppose n = 3. Find the joint p.m.f. of (X,Y), and the marginal p.m.f. of Y. 2. (A classification problem) Suppose n = 3. Find the conditional p.m.f. of X given Y = 3.

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Answer 1

(i)

Let X be the number of questions correctly answered by the student. Then, X ~ Binomial(n = 30, p = 0.7)

Probability that the student scores 100% on the test = P(X = 30) = 0.7³⁰ = 2.253934 x 10^-5

(ii)

Expected value of student's score = np = 30 * 0.7 = 21

(iii)

Variance in student's score = np(1-p) = 30 * 0.7 * (1 - 0.7) = 6.3

(iv)

60% of 30 = 18

70% of 30 = 21

Probability to score a credit = P(18 $\le$ X < 21) = P(X = 18) + P(X = 19) + P(X = 20)

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(iv)

Number of T answers are 13.

Expected number of times student correctly answers T = 13 * 0.7 = 9.1

(v)

Let T and F be the number of times student correctly answered T and F questions.

Then,

T ~ Binomial(n = 13, p = 0.7)

F ~ Binomial(n = 17, p = 0.7)

Probability that the student correctly answers T exactly 6 times and correctly answers F exactly 8 times

= P(T = 6) * P(F = 8)

(13 (171 0.76 * 0.313-6* 10.78 0.317-8 6

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