# Probability Problem...help needed!

by amywilliams99
Tags: probability, problemhelp
 Sci Advisor P: 6,104 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.
P: 4
 Quote by mathman 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.
Can I ask why 1/12 for the probability of doesnt know and guesses right? Wouldnt it be 1/4, given that there are 4 possible multiple choice answers?

P: 3,014
Probability Problem...help needed!

 Quote by amywilliams99 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. 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??
Since answering each question is independent from the others, it is sufficient to calculate the probability p that the student will answer the question right. Then the required probability is given by the Bernoulli distribution:

$$P_{9}(12, p) = \left( \begin{array}{c}12 \\ 9 \end{array} \right) p^{9} q^{3}, \ p + q = 1$$

On to calculating p! This is solved by the formula for total probability under conditional probability. Namely, let event $$A$$ be: "The student knows the correct answer" and let $$\overline{A}$$ be: "The student does not know the correct answer". From what is given, we have:

$$P(A) = \frac{2}{3}, \ P(\overline{A}) = 1 - P(A) = \frac{1}{3}$$

Then, we have $$B$$ be: "The student chooses the correct answer". We need the conditional probabilities $$P(B | A)$$ and $$P(B | \overline{A})$$ to calculate $$P(B) \equiv p$$. These are derived using logic and what is given:

$$P(B | A) = 1$$

because the student will definitely choose the correct answer if he knows the correct answer (event $$A$$).

$$P( B | \overline{A}) = \frac{1}{4}$$

because the student chooses at random out of 4 choices when he does not know the correct answer (event $$\overline{A}$$). Then, by the formula for the total probability:

$$p \equiv P(B) = P(A) P(B | A) + P(\overline{A}) P(B | \overline{A}) = \frac{2}{3} \cdot 1 + \frac{1}{3} \cdot \frac{1}{4} = \frac{2}{3} + \frac{1}{12} = \frac{9}{12} = \frac{3}{4}$$

All you need to do is substitute in the other numbers and compute the result. Good luck! :)
 P: 4 Thank you SO much, I understand!!

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