There are 3 questions in a multiple choice test, each question has four choices:

Question

There are 3 questions in a multiple choice test, each question has four choices: A to D. Each correct answer gets 1 point and each wrong answer deducts 1/3 point.

a) find the joint probability of having zero correct, 1 correct, 2 correct, 3 correct. And show that a random strategy will end up with zero points.

b) With the random strategy but one wrong answer is dropped (but keep all correct ones), what is the average points of such a test?

c) If one question is answered correctly, but remaining two are answerd with random strategy, what will be the average points earned?

Explanation / Answer

a) P(correct answer) = 1/4

P(wrong answer) = 3/4

P(zero correct) = (3/4)*(3/4)*(3/4) = 27/64

P(one correct) = select one correct answer out of three = (3C1)*(1/4)*(3/4)*(3/4) = 27/64

P(two correct) = select two correct answer out of three = (3C2)*(1/4)*(1/4)*(3/4) = 9/64

P(all correct) = (1/4)*(1/4)*(1/4) = 1/64

E(random strategy) = -1*P(0 correct) + (1/3)*P(1 correct) + (5/3)*P(2 correct) + 3*P(3 correct)

E(random strategy) = -1*(27/64) + (1/3)*(27/64) + (5/3)*(9/64) + 3*(1/64)

= 0

b)

In this case, scores associated with each probability changes.

E(random strategy) = (-2/3)*P(0 correct) + (2/3)*P(1 correct) + (2)*P(2 correct) + 3*P(3 correct)

E(random strategy) = (-2/3)*(27/64) + (2/3)*(27/64) + (2)*(9/64) + 3*(1/64)

= 21/64

c) In this case we have one question correct and we guess for remaining two

E(random strategy) = 1 + (-2/3)*P(0 of 2 correct) + (2/3)*P(1 of 2 correct) + (2)*P(2 of 2 correct)

E(random strategy) = 1 + (-2/3)*(3/4*3/4) + (2/3)*(2C1*1/4*3/4) + (2)*(1/4*1/4)

= 1 + (-2/16) + (2/16)

= 1

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