To mange this question mathematically we're going to need to assume some things. If something I'm going to assume is not what you intended then i apologize. We need to assume that it's not the same border checker. (whoever rejects the cigarettes). If it is, then its very probable they will be rejected. Anyways we'll assume its a binomial random variable with a succes chance of .94 and a failure rate of .06. Then, what you're asking, what is the probability of success at the border assuming that it has failed once before. That is, of those that failed once, what percentage will pass. Using bayes theorem, P(AnB)/P(B). Where P(AnB) is the probability of a fail then a pass, or p*q, .94*.06 or .0564. Then we need to divide that by .06 since that is our new universe, since we've assumed that you've failed once already. As you can see, it is again .94. The lesson learned. If you flip a coin over and over, your chance of getting heads or tails always remains the same. The same lesson can be applied to this. Instead of a coin with equal probabilities of heads or tails, your situation is a coin with a 94% chance of heads and a 6% chance of fail. Although, you may want to question the source that said 6% chance of failure, because they may be misleading. Also, you may now become flagged since you've had a package rejected. I'm not sure how that sort of thing works.