Question 1007053
i believe this will be equal to probability of getting exactly 5 of the numbers correct plus the probability of getting exactly 6 of the numbers correct.


the general formula is:


p(x) = p^x * q^(n-x) * nCx


nCx = n! / ((n-x)! * x!)


p = 1/48
q = 1 - 1/48 = 47/48


in your problem:


p(5) = p^5 * q^1 * 6C5
p(6) = p^6 * q^0 * 6C6


trsnalate to numbers and you get:


p(5) = (1/48)^5 * (47/48)^1 * 6 = 2.305694078 * 10^(-8)
p(6) = (1/48)^6 * (47/48)^0 * 1 = 8.17622013 * 10^(-11)


add them up and you get p(5 or 6) = 2.313870298 * 10^-8


that is equivalent to .00000002313870298


round that to 7 decimal places and you get .0000000 = 0


here's the printout of the excel worksheet where i did all the calculations.


the total probbility equals 1 as it should.


the probability of getting 5 right or 6 right is the sum of p(x) for x = 5 and 6.


the probability is 0 when rounded to 7 decimal places.


<img src = "http://theo.x10hosting.com/2015/120101.jpg" alt="$$$" </>