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A long probability question

So, my boyfriend ordered cigarettes from across the border and was told there would be a 6% chance it would be rejected at the border.  Unfortunately for his addiction, it was indeed rejected at the border.
My question is, at this point, does he still have a 6% chance of getting rejected at the border, or is the probability substantially lower considering he was already rejected?

2 Answers by Expert Tutors

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Kenneth G. | Experienced Tutor of Mathematics and StatisticsExperienced Tutor of Mathematics and Sta...
The two crossings are independent provided the boarder guards don't recognize the person taking the cigarettes across the boarder, so there should be no effect -  unless...
If your boyfriend takes the cigarettes across the boarder himself, and the boarder guards took his picture the first time or have a record of him or recall him crossing the boarder illegally with cigarettes previously, then they may profile him, giving a 100% chance of rejection at the boarder on the second attempt.
Drake O. | Solid Mathematical Foundations and Advanced Mathematics TutoringSolid Mathematical Foundations and Advan...
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.