text: A good analogy here is flipping a fair coin.  So like if you have flipped a coin 5 times and each time came up heads.  What is the probability that the 6th time will be heads as well?  \n\nAnswer: the same as it was the 1st, 2nd, 3rd... 50/50.  Each event is independent of every other event, so the probability is always the same for each individual event.  \n\nNow, if you flip 6 coins simultaneously, and you want to know what the probability of getting at least one head then you use the Binomial Distribution formula:\n\n  b(k;n,p) = (n choose k)(p^k)(1-p)^(n-k)\n\nWhere n = the total number of possible outcomes\nand k = the amount of desired outcomes\nand p = the probability of k occurring\nand 1 - p = the probability of k not occurring\n\nNote: "n choose k" is found using the general formula (n!/[(n-k)!k!]\nwhich is like...well say n =5 then n! is 5x4x3x2x1 (in the numerator) and say k = 3 then the denominator is (2x1)(3x2x1).  \n\nIs that clear as mud?