the probability of a 5/1 chance occuring at least once in five trials i come up with .598 5/6 to the fifth power how would you compute the probability of a 4k/1 occurring at least once in 4k trials thank you Harold Thomas
Your answer is incorrect. You have computed the probability of exactly 5 occurrences in five trials. The answer is a simple binomial probability. In general let p be the probability of success (in your case p=5/6). Then the probability of at least 1 success in n trials (in your case n=5) is 1- probability of zero success. This is 1-(1-p)^n, in your case 1-(1-5/6)^5=1-(1/6)^5=1-(1/7776)=1-0.00013=.99987. For the other case you mention p=4k/(4k+1) so the result is 1-(1-[4k/(4k+1)])^4k for k=1 it is .984 for k>=2 it is bigger than .999.
You are right about the limit but not the answer to his problem. It should be 1-(1/6)^5 which is a lot bigger than .598! It is 1- the probability that it never comes up in 5 trials not 1-the probability that it come up all 5 times!