Question 1152675
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p = 0.8 = probabiliity of success = probability of flight on time
n = 14 = sample size


Either the flight is on time (success) or it is not on time (failure). Let's assume, for the sake of simplicity, that each flight is independent of any other flight. Based on that, this means we have a binomial problem. 


We will use the binomial PDF function
{{{B(x) = ((n!)/(x!(n-x)!))*(p^x)*(1-p)^(n-x)}}}
to compute individual binomial probabilities. 
The expression {{{(n!)/(x!(n-x)!)}}} represents the binomial coefficient. It can be found alternatively through Pascal's Triangle. The exclamation marks represent factorials.


Plug in n = 14 and p = 0.8 to get
{{{B(x) = ((n!)/(x!(n-x)!))*(p^x)*(1-p)^(n-x)}}}


{{{B(x) = ((14!)/(x!(14-x)!))*(0.8^x)*(1-0.8)^(14-x)}}}


From here we need to compute B(x) for x = 12, 13 and 14. 
Then we'll add up those results.


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Plug in x = 12


{{{B(x) = ((14!)/(x!(14-x)!))*(0.8^x)*(1-0.8)^(14-x)}}}


{{{B(12) = ((14!)/(12!(14-12)!))*(0.8^12)*(1-0.8)^(14-12)}}}


{{{B(12) = 0.25014}}} This represents the approximate probability of exactly 12 flights are on time. 


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Plug in x = 13


{{{B(x) = ((14!)/(x!(14-x)!))*(0.8^x)*(1-0.8)^(14-x)}}}


{{{B(x) = ((14!)/(13!(14-13)!))*(0.8^13)*(1-0.8)^(14-13)}}}


{{{B(13) = 0.15393}}} This represents the approximate probability of exactly 13 flights are on time. 


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Plug in x = 14


{{{B(x) = ((14!)/(x!(14-x)!))*(0.8^x)*(1-0.8)^(14-x)}}}


{{{B(x) = ((14!)/(14!(14-14)!))*(0.8^14)*(1-0.8)^(14-14)}}}


{{{B(14) = 0.04398}}} This represents the approximate probability of exactly 14 flights are on time. 


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Add up the results to find the probability that x is 12 or larger.


{{{B(x >= 12) = B(12) + B(13) + B(14)}}}


{{{B(x >= 12) = 0.25014 + 0.15393 + 0.04398}}}


{{{B(x >= 12) = 0.44805}}}


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<font size=4 color=red>Answer: approximately 0.44805</font>
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