Question 905632
the 12/6 should be 12 C 6. This is referring to the combination formula


<a href="http://www.mathwords.com/c/combination_formula.htm">http://www.mathwords.com/c/combination_formula.htm</a>


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n C r = (n!)/(r!(n-r)!)


12 C 6 = (12!)/(6!*(12-6)!)


12 C 6 = (12!)/(6!*6!)


12 C 6 = (12*11*10*9*8*7*6!)/(6!*6!)


12 C 6 = (12*11*10*9*8*7)/(6!)


12 C 6 = (12*11*10*9*8*7)/(6*5*4*3*2*1)


12 C 6 = (665280)/(720)


12 C 6 = 924

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So you should have this


(12 C 6)*(0.71)^6*(0.29)^6 


(924)*(0.71)^6*(0.29)^6 


0.0704060615439


Round that to 4 decimal places as instructed to get 0.0704


Therefore, P(Exactly 6 pass the exam) = <font color="red">0.0704</font> (approximately)


This is roughly a 7.04% chance.


Here is a handy binomial distribution stats calculator


<a href="http://stattrek.com/online-calculator/binomial.aspx">http://stattrek.com/online-calculator/binomial.aspx</a>