Question 368252
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You really cannot calculate this probability in any real sense.  Just because you know that the student knows how to do 13 of the 17 problems correctly tells you nothing about how the student is likely to perform on the other four problems.  If you add the condition that if one of the remaining 4 problems appears on the exam that the student is guaranteed to answer incorrectly, then proceed as follows:


The probability of *[tex \Large k] successes in *[tex \Large n] trials where *[tex \Large p] is the probability of success on any given trial is given by:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ P_n(k,p)\ =\ \left(n\cr k\right\)p^k\left(1\,-\,p\right)^{n\,-\,k}]


Where *[tex \LARGE \left(n\cr k\right\)] is the number of combinations of *[tex \Large n] things taken *[tex \Large k] at a time and is calculated by *[tex \Large \frac{n!}{k!(n\,-\,k)!}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ P_6\left(6,\frac{13}{17}\right)\ =\ \left(6\cr 6\right\)\left(\frac{13}{17}\right)^6\left(\frac{4}{17}\right)^0]


Note that *[tex \LARGE \left(n\cr n\right\)\ =\ 1\ \forall\ n\ \in\ \mathbb{N}] and *[tex \LARGE x^0\ =\ 1\ \forall\ x\ \in\ \mathbb{R}]


Hence:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ P_6\left(6,\frac{13}{17}\right)\ =\ \left(\frac{13}{17}\right)^6]


You can do your own arithmetic.


John
*[tex \LARGE e^{i\pi} + 1 = 0]
My calculator said it, I believe it, that settles it
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