<PRE><B>Question 1043335</B>
For typographical errors, it is a Poisson distribution with small number of errors that can theoretically be infinite, are considered to be independent, integer amounts, and occurrence is proportional to interval length.
for lambda=4, the probability of x events&#39; occurring is e^(-lambda)*lambda^x/x!
probability of 6 is e^(-4)4^6/6!
This is 0.0183*4096/720=0.104
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probability of 5 is e^(-4)4^5/5!=0.1563
probability of 4 is (e^(-4)4^4/4!=0.1954
probability of 3 is e^-4)4^3/3!=0.1954
probability of 2 is e^(-4)4^2/2=0.1465
probability of 1 is e^(-4)*4=0.0733
probability of 0 is e^(-4)=0.0183
At most 5 is the sum, or 0..785
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At most 6 is 0.785+ probability of 6, or 0.104 from above, and that is 0.889
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More than 6 is the complement of 1-0.889 or 0.111.
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