This is the security check problem from the quality module practice problems. This problem is about travelers leaving the EU who face security screening at up to four different test points. These test points are as follows. First, passengers must obtain a boarding pass. Airlines recognize 50% of security violations here. Then passengers are screened by the German border patrol. They recognize 90% of security violations. Third, the airline screens passengers at the gate. This catches 30% of security violations. Finally 5% of passengers are chosen for screening just before entering the plane. This process catches 99% of security violations. The first question asks, what is the probability that a passenger who is guilty of a security violation makes it past the gate? That is, what is the probability that a guilty passenger makes it past the first three test points undetected? This is the probability that the passenger obtains a boarding pass which is 1 minus 0.5. Times the probability that the passenger is not caught by border patrol, which is 1 minus 0.9. Times the probability that the passenger is not caught at the gate, which is 1 minus 0.3. This simplifies to 0.5 times 0.1 times 0.7, which equals 0.035, or 3.5%. The second question asks, what the probability a guilty passenger actually boards the plane is? This is the probability that the passenger makes it past the gate, times the probability that the passenger makes it past the fourth checkpoint. That is, 0.035 times 0.95 for the 95% chance the passenger is not selected for screening, plus 0.05 times 1 minus 0.99. For the 5% chance the passenger is selected for screening, times the probability the violation is caught. Notice that I have two sets of parenthesis here so we need to be mindful of order of operations. This simplifies to 0.035 times 0.95 plus 0.05 times 0.01, which is 0.0332675, or about 3.3%.