Question 485391
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Like many things in this life there is a hard way and an easy way to answer this question.


If 4 students take the test, then there are 5 situations that could possibly arise:


None pass, 1 passes, 2 pass, 3 pass, or all 4 pass.


Note for future reference that since these five outcomes are inclusive of all possible outcomes, the sum of the probabilities of each must be 1, because it is a certainty that one of the outcomes will occur.


Now, your question asks for the probability that <i>at least one</i> student passes.  Looking at it in a straight-forward way, that would be the sum of the probabilities that <i>exactly</i> 1 passes plus <i>exactly</i> 2 pass plus...and so on.


Since "Pass/Fail" is an either/or outcome, and the probability of success for any given instance of one student taking one test is given as 60%, we know that the probability should be calculated using the binomial distribution.


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\)\left(p\right)^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)!}]


But you would need to perform this calculation four times and then sum the results.  First you would need to calculate:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ P_4(1,0.6)\ =\ \left(4\cr 1\right\)\left(0.6\right)^1\left(0.4\right)^{3}]


To get the probability for exactly 1, and then you would have to do exactly 2, and so on.  In summary:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ P_4(\geq{1},0.6)\ =\ \sum_{i=1}^4\left(4\cr i\right\)\left(0.6\right)^i\left(0.4\right)^{n\,-\,i}]


But fortunately, there is a much simpler way.  From the point of view of your question, there are only two possible outcomes.  Either 1 or more pass, or nobody passes.  So if you take the probability that nobody passes and subtract that from 1, you get the probability that at least 1 passes.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ P_4(1,0.6)\ =\ 1\ -\ P_4(0,0.6)\ =\ 1\ -\ \left(4\cr 0\right\)\left(0.6\right)^0\left(0.4\right)^{4}]


Since we know that *[tex \Large n] pick 0 is 1 for all positive integers *[tex \Large n] and that *[tex \Large a^0\ =\ 1] for all real numbers *[tex \Large a],


The above reduces to:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 1\ -\ \left(0.4\right)^4]


The arithmetic is yours to do.


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