Ther's a lot of security around here due to the DNC. The public transportation system security personnel will be randomly searching the person and bags of commuters. Question: Assuming that no racial profiling is allowed; the search is thorough; there are 100,000 passengers, 10 of whom are bomb carrying terrorists; and every tenth person is searched: What is the probability of at least one of the terrorists getting through Security?
Posts: 625 | Location: Boston | Registered: 06-13-02
I just realized that if all 10 terrorists lined up at once, that the chances of not only one, but 9 of them, would get through, is 100%! I don't like my odds! That's why I'm going to refuse to have my privacy invaded, because the bad guy is going to get through anyway. For my questuion, however, assume that the 10 bad guys are randomly located 1 amongst each 10,000 passengers. *************************************************** 07-27-04, 09:20 AM DvdGStwrt Unfortunately I do not know how to start figuring up the statistics on this. I would imagine that the statistics would be somewhere around 1 out of 50,000 of being caught - Thats without doing any math on it, a guess.
Boston, hm. Democratic Convention - They still have that "freedom of Speech" area up - the Caged area near the new overpass? More logical thinking there. Roll Eyes ******** 07-27-04, 03:22 PM FlyingHellfish The way the question is posed, the probability of at least 1 of the 10 terrorists getting through security is 65.13%
To show why, think of it this way: The probability of at least (let's designate this by P[at least 1] one getting through is the same as the probability of exactly one, P[1],getting through plus the probability of exactly two, P[2], getting through plus the probability of exactly three, P[3] and so on up to P[10]. Put in mathematical terms:
That pretty much covers every possibility, except for the probability that zero terrorists get through, P[0]. Since the sum of all possible outcomes is 1, then:
But we know that P[1] + P[2]....+ P[10] = P[at least 1], so substituting, we get:
P[0] + P[at least 1] = 1
or
P[at least 1] = 1 - P[0]
The reason we put it in terms of just P[0] instead of P[1] to P[10] is because P[0] is by far the easiest to calculate. As we said, P[0] is the probability that none get through. Since 1 out of every 10 people get searched, that means that 9 out of every 10 people don't, or 9,000 out of every 10,000. Since there is only one terrorist out of every 10,000 people, the chance that he is among the 9,000 that don't get searched is 90%. This is the same probability for all ten terrorists. So, the probability that zero of the ten terrorists, which was P[0], get searched is 90%10, or 0.910, which is 34.87%.
Since P[at least 1] = 1 - P[0], then the probability that at least 1 terrorist gets through is 1 - 0.3487, or 65.13%. ********* 07-27-04, 10:12 PM gerry Thanks, FH, that was a great solution, but I'm still confused, because it seems like if there was instead only one (not ten) terrorist amongst the 100,000, he'd have a better chance of getting through (90%) than the chances of at least one of ten getting through (65%) if there were 10 terrorists. Please explain this, if my logic is correct. ******** 07-28-04, 12:59 AM FlyingHellfish Yes, that is true. Think of it this way: if there is only one terrorist who has a 90% chance of getting through, that means that there is a 10% chance of not getting through. That means that security has a 10% chance of finding this one terrorist. Now compare this to playing the lottery. What if you could play in a lottery where each ticket had a 10% of winning? How would your odds change if you had two tickets instead of one? Naturally, they would increase, and three tickets would be better than two and so on. The same principle applies here. More people that security has to look out for equates to higher probability of apprehending at least 1 of them, which means that it's harder (lower probability) for the entire group of terrorists to get by. ******** 07-28-04, 07:45 AM gerry Yes, FH, there is a lower probability for the entire group to get by, but there is also a lower probability that at least one will get by. What I am saying is that if a terrorrist acts alone in the crowd, he has a 90% chance of getting through, whereas if he he acts with 9 others, there is only a 65% chance that he or someone else or all of the buggers will get through. It seems as though if some government was bent on doing harm, it would be better to send just one guy over rather than a bunch of them. I'm still missing something here, what is wrong with my reasoning? ********* 07-28-04, 04:58 PM FredPuli The chance of any one individual terrorist ( M Mouse say), him or herself, being caught there in a trap is always the same. Likewise the chance of his terrorist friend (D Duck)of being caught, perhaps by some decoy, is the same, quite regardless of how many others there are in the crowd.There could be terrorists S. White, and seven others too, but the chance of getting Grumpy is no different from the chance of getting Happy. Don't get Bashful : you can say , without being Dopey, sorry, dopey,that the more terrorists there are in the the crowd the more chance you have of catching a terrorist. By the end you may even have got Sleepy but the sad fact remains that, if only the one ever had and still has has a bomb concealed about them, only the explosion may show you what's up Doc.[See? It's Sneezy to work it out] ******** 07-28-04, 09:17 PM gerry Thanks, Fred, now I know the names of all the 7 dwarfs, but I still don't have a good answer to my question. How can Dopey have a 90% chance of getting through, if he acts alone in a group of 100,000, whereas if he acts with 9 other dwarfs in that group, the chances of he or one or more of his friends getting through is only 65%??? If this is really the case, a smart terrorist government would send just one guy over instead of 10, the odds of the lone terrorist creating havoc are better than one of the 10 creating havoc.I just don't understand it. ******** 07-29-04, 01:00 AM decal "So, the probability that zero of the ten terrorists, which was P[0], get searched is 90%10, or 0.910, which is 34.87%.
Since P[at least 1] = 1 - P[0], then the probability that at least 1 terrorist gets through is 1 - 0.3487, or 65.13%...."
That is to say, the probability of ALL TEN getting through is 65.13%, yes? The probability of ANY PARTICULAR one is still 90%, though--or am I reading this wrong? ******** 07-29-04, 02:03 AM FredPuli We need a bookmaker. Even I can work out that with one terrorist, Dopey, among ten people I've a one in ten chance of guessing which is him at my one guess. And if I have two terrorists, Dopey and Grumpy in the 10 I've a two in ten chance of picking a terrordwarf. If I guess right and pick a terrordwarf the bookmaker is not likely to think that my chances of finding the second are still one in ten Big Grin. I've reduced the field to nine having taken one out so it's now one in nine of finding the second. However it's a mistake that the layman might make, thinking that the odds are still the same and it might lead to his doing some puzzling but erroneous calculation that flies in the face of fact. If there were five terrorists in ten this belief could put him way out; hunting one named individual starts at one in ten but is down to one in six if he finds the other four straight off (or doesn't but removes or checks off those who are not the named terrorist as he goes)
Tell me, as a layman, is this the basis of the odd mathematical results between finding of one individual,in a group of five terrorists when all are in a fixed number of people but not getting the same apparent probability when only the one of them is a group of that size?
There's plainly some bad maths somewhere.
BTW a real bookmaker could tell me the odds on a forecast. Straight forecast: Dopey and Grumpy are in a field of ten. Pick out the two from the field of ten picking Grumpy on your very first guess and Dopey on your second. Reverse forecast: picking Dopey and Grumpy in the first two guesses, in either order, Dopey followed by Grumpy or Grumpy followed by Dopey; either 'forecast' winning.
Can you? ( The reverse forecast and its variants; of first three in any order, first four in any and so on are also known to him as are the odds on your guessing right again if you are lucky on your first two, first three etc Isn't it shaming to the rest of us that he can do this when simultaneously working with the different odds quoted for each horse in a race ? The old style settlers could do this with two or three horse forecasts in their head, no computers ) ******** 07-30-04, 08:27 AM FlyingHellfish
quote:....the probability that zero terrorists get through, P[0]...
quote:...the probability that zero of the ten terrorists, which was P[0], get searched...
My apologies...I should've known something was off with my answer, and if you read closely the two lines I quoted from my original post, you'll see where I got muddled Smile
Going back to the first definition of P[0], which is the probability that zero terrorists get through (or put differently, the probability that all ten get caught), we can still employ the exact same method. The chance for any one of them to get caught is 10% or 0.1. Hence, the chance for all ten of them to get caught is 10%10, or 0.0000001%. Going back to the original equation:
P[at least 1] = 1 - P[0]
we get
P[at least 1] = 1 - 0.000000001
or 99.9999999% chance that at least one gets through.
The 34.87% refers to the probability that all terrorists get through. Likewise, there's a 65.13% chance that at least one gets caught Smile
Sorry about the mistake.
This message has been edited. Last edited by: DorianGreyed,
Posts: 625 | Location: Boston | Registered: 06-13-02
