(New page: I think this tip could help solve this problem. The PMF for the questions he already knows is 0. Lets take the number of question he knows and named that variable m (instead of k for sim...) |
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Latest revision as of 01:31, 24 September 2008
I think this tip could help solve this problem. Please correct me if I'm wrong.
The PMF for the questions he already knows is 0.
Lets take the number of question he knows and named that variable m (instead of k for simplicity). For this set of questions the probability of getting it righ is p=1 (assuming he knows the answer to the question and is going to get it right). Then,
PMF = (n over k) * p^k * (1-p)^k ==> pmf function for a binomial R.V.
We know that p = 1 in this case, so the term (1-p) goes to 0 making the whole pmf equal to 0. So to calculate the distribution we can pretty much neglect the questions he already know (think of the distribution of this questions as a constant that won't affect the randomness of the outcome of the questions he guessed on).
The set of questions that he already know only make difference when calculating the expected value because they do alter the average.
