
#1
Apr1710, 03:38 PM

P: 4

A student is about to take a test, which consists of 12 multiple choice
questions. Each question has 4 possible answers, of which only one is correct. The student will answer each question independently. For each question, there is a probability 2/3 that he knows the correct answer; otherwise, he picks an answer at random. (a) Compute the probability that he will get exactly 9 questions right. (b) compute the probability that his 12th answer will be his 9th correct answer (c) Compute the probability that his 6th answer will be the 5th correct answer and his 12th answer will be the 9th correct answer. Please help. For a, I think its binomial distribution, but I am having trouble with the probability, Is it simply 2/3 or must you factor in the probability for guessing. ie  probability is 2/3 +1/4 for the correct answer?? 



#2
Apr1710, 04:12 PM

Sci Advisor
P: 5,935

On any given question the probability that he gets the correct answer is 2/3 (knows) + 1/12 (doesn't know and guesses right) = 3/4. Use binomial to work out the probabilities.




#3
Apr1710, 04:24 PM

P: 4





#4
Apr1710, 04:48 PM

P: 3,015

Probability Problem...help needed![tex] P_{9}(12, p) = \left( \begin{array}{c}12 \\ 9 \end{array} \right) p^{9} q^{3}, \ p + q = 1 [/tex] On to calculating p! This is solved by the formula for total probability under conditional probability. Namely, let event [tex]A[/tex] be: "The student knows the correct answer" and let [tex]\overline{A}[/tex] be: "The student does not know the correct answer". From what is given, we have: [tex] P(A) = \frac{2}{3}, \ P(\overline{A}) = 1  P(A) = \frac{1}{3} [/tex] Then, we have [tex]B[/tex] be: "The student chooses the correct answer". We need the conditional probabilities [tex]P(B  A)[/tex] and [tex]P(B  \overline{A})[/tex] to calculate [tex]P(B) \equiv p[/tex]. These are derived using logic and what is given: [tex] P(B  A) = 1 [/tex] because the student will definitely choose the correct answer if he knows the correct answer (event [tex]A[/tex]). [tex] P( B  \overline{A}) = \frac{1}{4} [/tex] because the student chooses at random out of 4 choices when he does not know the correct answer (event [tex]\overline{A}[/tex]). Then, by the formula for the total probability: [tex] p \equiv P(B) = P(A) P(B  A) + P(\overline{A}) P(B  \overline{A}) = \frac{2}{3} \cdot 1 + \frac{1}{3} \cdot \frac{1}{4} = \frac{2}{3} + \frac{1}{12} = \frac{9}{12} = \frac{3}{4} [/tex] All you need to do is substitute in the other numbers and compute the result. Good luck! :) 



#5
Apr1710, 04:55 PM

P: 4

Thank you SO much, I understand!!



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