# Okay math people...

#### Bacillus anthracis

##### Member
Say:

Each flight that comes into an airport carries 100 passengers.

20% of the people on each flight will be randomly selected by a computer searched once they de-board the plane. Again, the selection is utterly arbitrary.

On each plane, there is one terrorist/child porn smuggler/drug runner/Bad Guy (it doesn't really matter). But we have no idea who that bad guy is.

What are the chances (either by odds or percentage) of the terrorist being caught?

What if there are two terrorists?

Thanks in advance. If I ever see you IRL I'll buy you a beer or a coffee or whatever your poison is. I'm not buying anyone a pint of melted gold.

#### Kharakov

##### Quantum Hot Dog
0. Tom Sawyer works for the TSA...

#### EricK

##### Senior Member
If there is one bad guy and you check 20% of the people, the chances you catch him (assuming you are guaranteed to realise he is the bad guy once you search him) is 20%. And the chance he gets through is 80%

If there are two bad guys, then
Both get through 63.84% of the time
One gets through and you catch one 32.32% of the time
You catch both 3.84% of the time

If there are 3 then you catch:
0 50.81% of the time
1 39.08% off the time
2 9.40% of the time
3 0.71% of the time

At least that's the answer if I have remembered how to do these sums. It is a good sign that my answers add up to 100% in each case

#### Jimmy Higgins

##### Contributor
0%

What is the point of getting the terrorist after they have boarded?!

#### beero1000

##### Veteran Member
If there is one bad guy and you check 20% of the people, the chances you catch him (assuming you are guaranteed to realise he is the bad guy once you search him) is 20%. And the chance he gets through is 80%

If there are two bad guys, then
Both get through 63.84% of the time
One gets through and you catch one 32.32% of the time
You catch both 3.84% of the time

If there are 3 then you catch:
0 50.81% of the time
1 39.08% off the time
2 9.40% of the time
3 0.71% of the time

At least that's the answer if I have remembered how to do these sums. It is a good sign that my answers add up to 100% in each case

For the record - this is correct.

You are looking at a hypergeometric distribution with parameters m varying 1,2,3, n = 20, and N = 100.

Each probability is then given by P(X = k) = C(N-m,n-k)*C(m,k)/C(N,n), where C(a,b) represents binomial coefficients.

#### NobleSavage

##### Veteran Member
What are the chances (either by odds or percentage) of the terrorist being caught?
Very slim. I know this isn't what you are asking, but:

An internal investigation of the Transportation Security Administration (TSA) found that undercover investigators were able to smuggle fake explosives and weapons through checkpoints in 95 percent of trials, which they conducted at dozens of America’s busiest airports. Officials did not disclose when the testing took place, other than to say it ended recently.

http://www.newsweek.com/airport-security-fails-detect-95-fake-explosives-weapons-337932