Probability: On a multiple choice exam with 4 choices for each question,?

Question

a student either knows the answer to a question or marks it random. If the probability that the student knows the answer is 2/3, what is the probability that an answer that was marked correctly was not marked randomly?

Please try to answer using Bayes' forumla and show steps.

Answer 1

The English used in wording probability problems can easily give the wrong impression of what is being asked.

Lets first list the probability values that we are given or that are immediately implied:

A - event student knows the answer, P(A) = 2/3
A'- event student does not know the answer, P(A') = 1/3

(note event that student does not know the answer is the same as the event that the answer was chosen randomly)

C - event the question was answered correctly

Note that P(A) = P(A^C) = 2/3, maybe a twist over the regular bayes formula thinking since there is a dependence working here.
I don't recall seeing a dependence like this before, but that shows how little I have studied the subject.


P(C) = P(A ^ C) + P(A' ^ C) = 2/3 + (1/3)*(1/4) = 9/12

We want to know the probability that given the answer was correct, it had not been chosen at random (meaning the student actually knew the answer.)

P(A|C) = P(A^C)/P(C) = (2/3)/(9/12) = 8/9

I could have easily made an error for a variety of reasons, please check my work, maybe the last equation can be put in Bayes form. This is all I know at present.

Answer 2

I won't use any preassigned method, that is your job to understand use. As for the solution here it is:
A correct solution can come from two sources, which is that the student knew the answer and got it right which would have happened with a probability of 2/3 and the guy did not knew but guessed it right which would happen with a probability of
1/3 x1/4=1/12.
So the probability of knowing the answer given the correctness of the answer is
2/3 (prob of knowing)
----------------------------
2/3+1/12 (prob of corr)
If you simplify, you get 8/9.

Answer 3

a) P= a million/4*a million/4*a million/4*a million/4*a million/4 = a million/(4^5) b) Use binomial distribution (cdf not pdf) and use n=5, p=.2 and x=3 to get P(x<=3) then subtract that from a million to get P(x>=4) c) P = 3/4*3/4*3/4*3/4*3/4 d) Use binomial distribution cdf and use n=5 p=.2 and x=2 to get P(x<=2)

Answer 4

Use your instinct if u are not for sure