(a) [4 marks] A student answers a multiple-choice examination question that offersfour possible answers. Suppose that the probability that he knows the answer tothe question is 0.8 and the probability that he guesses is 0.2. Assume that if thestudent guesses, the probability of selecting the correct answer is 0.25. If the studentcorrectly answers a question, find the probability that he really knew the correctanswer.(b) [5 marks] Suppose the student takes an examination with 3 multiple choice ques-tions. Let X denote the number of questions he gets correct. Assuming that thestudent answers each question independently, determine the probability distributionof X and calculate the expected value of X.(c) [10 marks] Suppose the student takes an examination with 2 multiple choice ques-tions. But this time, he gains confidence every time he knows the answer to a ques-tion. Specifically, if he knows the answer to the current question, then for the nextquestion the probability that he knows the answer becomes 0.9 and the probabilitythat he guesses becomes 0.1. If he does not know the answer to the current ques-tion, then for the next question the probability that he knows the answer and theprobability that he guesses remain 0.8 and 0.2, respectively. Again letting X denotethe number of questions he gets correct, determine the probability distribution of Xand calculate the expected value of X. We can assume the probabilities of knowingthe answer and guessing for the first question are 0.8 and 0.2, respectively.
Question
(a) [4 marks] A student answers a multiple-choice examination question that offersfour possible answers. Suppose that the probability that he knows the answer tothe question is 0.8 and the probability that he guesses is 0.2. Assume that if thestudent guesses, the probability of selecting the correct answer is 0.25. If the studentcorrectly answers a question, find the probability that he really knew the correctanswer.(b) [5 marks] Suppose the student takes an examination with 3 multiple choice ques-tions. Let X denote the number of questions he gets correct. Assuming that thestudent answers each question independently, determine the probability distributionof X and calculate the expected value of X.(c) [10 marks] Suppose the student takes an examination with 2 multiple choice ques-tions. But this time, he gains confidence every time he knows the answer to a ques-tion. Specifically, if he knows the answer to the current question, then for the nextquestion the probability that he knows the answer becomes 0.9 and the probabilitythat he guesses becomes 0.1. If he does not know the answer to the current ques-tion, then for the next question the probability that he knows the answer and theprobability that he guesses remain 0.8 and 0.2, respectively. Again letting X denotethe number of questions he gets correct, determine the probability distribution of Xand calculate the expected value of X. We can assume the probabilities of knowingthe answer and guessing for the first question are 0.8 and 0.2, respectively.
Solution
(a) We can use Bayes' theorem to solve this problem. The probability that the student knew the answer given that he answered correctly is the probability that he knows the answer and answers correctly divided by the probability that he answers correctly.
The probability that he knows the answer and answers correctly is simply the probability that he knows the answer, which is 0.8.
The probability that he answers correctly is the sum of the probability that he knows the answer and answers correctly and the probability that he guesses and answers correctly. The probability that he guesses and answers correctly is the probability that he guesses (0.2) times the probability that he guesses correctly (0.25), which is 0.05.
So, the probability that he answers correctly is 0.8 + 0.05 = 0.85.
Therefore, the probability that he knew the answer given that he answered correctly is 0.8 / 0.85 = 0.941.
(b) The probability distribution of X can be determined using the binomial distribution. The probability of getting x correct answers is (3 choose x) * (0.85)^x * (0.15)^(3-x), where (3 choose x) is the number of ways to choose x correct answers from 3 questions.
The expected value of X is the sum of x times the probability of getting x correct answers, which is 3 * 0.85 = 2.55.
(c) This is a more complex problem because the probabilities change depending on whether the student knew the answer to the previous question. We can solve it by considering all possible sequences of knowing and guessing.
The probability distribution of X is:
P(X=0) = 0.2 * 0.75 * 0.2 * 0.75 = 0.0225 (he guesses and gets both wrong) P(X=1) = 0.2 * 0.75 * 0.8 + 0.8 * 0.2 * 0.75 = 0.255 (he guesses and gets one right and knows and gets one wrong) P(X=2) = 0.8 * 0.8 + 0.8 * 0.2 * 0.25 + 0.2 * 0.75 * 0.9 = 0.7225 (he knows and gets both right or he guesses and gets one right and knows and gets one right)
The expected value of X is 0 * P(X=0) + 1 * P(X=1) + 2 * P(X=2) = 0 + 0.255 + 2 * 0.7225 = 1.7.
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