Two bags are randomly selected from a flight at the airport to be searched, one at a time, then sent through after being searched (this means: without return). Five smugglers are watching this and they each have three bags with contraband. There are 20 bags on the flight. What is the probability that they will get away? (Answer in lowest terms).  Choices
A) 5/75 B) 3/12 C) 5/76 D) 2/19 E) 1/2
The flight has 20 bags total. There are (5 smugglers)(3 bags) = 15 bags with contraband, and   20 - 15= 
<gadget id="calculator">  20 - 15</gadget>
<output>5</output>
5 with no contraband, so the probability of getting a bag with no contraband is P(N)= 5/20 on the first selected bag.   Because selection is (without return), the P(N) has changed for following selections. Now, there are 19 bags, but the number bags with contraband stay the same; since, we are only calculating for success as no bag with contraband... and so on.... Therefore, conditional probability! The probability that event B occurs, given that event A has already occurred is:   Formula = P(A\B) = P(A and B)/P(B).   2 bags are selected; so,   P(A\B) = (5/20)(5/19) = 25/380 = 5/76   Answer: 5/76
<result>C</result>
Two bags are randomly selected from a flight at the airport to be searched, one at a time, then sent through after being searched (this means: without return). Five smugglers are watching this and they each have three bags with contraband. There are 20 bags on the flight. What is the probability that they will get away? (Answer in lowest terms).