{"id":6832,"date":"2017-01-24T16:37:57","date_gmt":"2017-01-24T16:37:57","guid":{"rendered":"http:\/\/www.rationalargumentator.com\/index\/?p=6832"},"modified":"2017-01-24T16:37:57","modified_gmt":"2017-01-24T16:37:57","slug":"terrorist-probability","status":"publish","type":"post","link":"https:\/\/www.rationalargumentator.com\/index\/blog\/2017\/01\/terrorist-probability\/","title":{"rendered":"What Are the Chances That a Muslim Is a Terrorist? – Article by Sanford Ikeda"},"content":{"rendered":"
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\"The<\/a>Sanford Ikeda<\/span>
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It\u2019s flu season and for the past two days you\u2019ve had a headache and sore throat. You learn that 90% of people who actually have the flu also have those symptoms, which makes you worry. \u00a0Does that mean the chances of your having the flu is 90%? \u00a0In other words, if there\u2019s a 90% chance of having a headache and sore throat given that you have the flu, does that mean there\u2019s a 90% chance having the flu given that you have a headache and sore throat?We can use symbols to express this question as follows: Pr(Flu | Symptoms) = Pr(Symptoms | Flu) = 90%?<\/p>\n

The answer is no. Why?<\/p>\n

If you think about it you\u2019ll realize that there are other things besides the flu that can give you a combination of a headache and sore throat, such as a cold or an allergy, so that having those symptoms is certainly not the same thing as having the flu. \u00a0Similarly, while fire produces smoke, the old saying that \u201cwhere there\u2019s smoke there\u2019s fire\u201d is wrong because it\u2019s quite possible to produce smoke without fire.<\/p>\n

Fortunately, there\u2019s a nice way to account for this.<\/p>\n

How Bayes\u2019 Theorem Works<\/strong><\/p>\n

Suppose you learn that, in addition to Pr(Symptoms | Flu) = 90%, that the probability of a randomly chosen person having a headache and sore throat this season, regardless of the cause, is 10% \u2013 i.e. Pr(Symptoms) = 10% \u2013 and that only one person in 100 will get the flu this season \u2013 i.e. Pr(Flu) = 1%. \u00a0How does this information help?<\/p>\n

Again, what we want to know are the chances of having the flu, given these symptoms Pr(Flu | Symptom). \u00a0To find that we\u2019ll need to know first the probability of having those symptoms if we have the flu (90%) times the probability of having the flu (1%). \u00a0In other words, there\u2019s a 90% chance of having those symptoms if in fact we do have the flu, and the chances of having the flu is only 1%. That means Pr(Symptoms | Flu) x Pr(Flu) = 0.90 x 0.01 = 0.009 or 0.9% or a bit less than one chance in 100.<\/p>\n

Finally, we need to divide that result by the probability of having a headache and sore throat regardless of the cause Pr(Symptoms), which is 10% or 0.10, because we need to know if your headache and sore throat are flu Symptoms out of all headache-and-sore symptoms that have occurred.<\/p>\n

So, putting it all together, the answer to the question, \u201cWhat is the probability that your Symptoms are caused by the Flu?\u201d is as follows:<\/p>\n

Pr(Flu | Symptoms) = [Pr(Symptoms | Flu) x Pr(Flu)] \u00f7 Pr(Symptoms) = 0.90 x 0.01 \u00f7 0.10 = 0.09 or 9%.<\/p>\n

So if you have a headache and sore throat there\u2019s only a 9% chance, not 90%, that you have the flu, which I\u2019m sure will come as a relief!<\/p>\n

This particular approach to calculating \u201cconditional probabilities\u201d is called Bayes\u2019 Theorem<\/a>, after Thomas Bayes, the 18th century Presbyterian minister who came up with it. The example above is one that I got out this wonderful little book<\/a>.<\/p>\n

Muslims and Terrorism<\/strong><\/p>\n

Now, according to some sources (here<\/a> and here),<\/a> 10% of Terrorists are Muslim. Does this mean that there\u2019s a 10% chance that a Muslim person you meet at random is a terrorist? \u00a0Again, the answer is emphatically no.<\/p>\n

To see why, let\u2019s apply Bayes’ theorem to the question, “What is the probability that a Muslim person is a Terrorist?” Or, stated more formally, “What is the probability that a person is a Terrorist, given that she is a Muslim?” or Pr(Terrorist | Muslim)?<\/p>\n

Let\u2019s calculate this the same way we did for the flu using some sources that I Googled and that appeared to be reliable. \u00a0I haven\u2019t done a thorough search, however, so I won’t claim my result here to be anything but a ballpark figure.<\/p>\n

So I want to find Pr(Terrorist | Muslim), which according to Bayes’ Theorem is equal to\u2026<\/p>\n

1) Pr(Muslim | Terrorist): \u00a0The probability that a person is a Muslim given that she\u2019s a Terrorist is about 10% according to the sources I cited above, which report that around 90% of Terrorists are Non-Muslims.<\/p>\n

Multiplied by…<\/p>\n

2) Pr(Terrorist): \u00a0The probability that someone in the United States is a Terrorist of any kind, which I calculated first by taking the total number of known terrorist incidents in the U.S. back through 2000 which I tallied as 121 from this source<\/a>\u00a0 and as 49 from this source<\/a>. At the risk of over-stating the incidence of terrorism, I took the higher figure and rounded it to 120. \u00a0Next, I multiplied this times 10 under the assumption that on average 10 persons lent material support for each terrorist act (which may be high), and then multiplied that result by 5 under the assumption that only one-in-five planned attacks are actually carried out (which may be low). \u00a0(I just made up these multipliers because the data are hard to find and these numbers seem to be at the higher and lower ends of what is likely the case and I\u2019m trying to make the connection as strong as I can; but I\u2019m certainly willing to entertain evidence showing different numbers.) \u00a0This equals 6,000 Terrorists in America between 2000 and 2016, which assumes that no person participated in more than one terrorist attempt (not likely) and that all these persons were active terrorists in the U.S. during those 17 years (not likely), all of which means 6,000 is probably an over-estimate of the number of Terrorists.<\/p>\n

If we then divide 6,000 by 300 million people in the U.S. during this period (again, I\u2019ll over-state the probability by not counting tourists and visitors) that gives us a Pr(Terrorist) = 0.00002 or 0.002% or 2 chances out of a hundred-thousand.<\/p>\n

Now, divide this by…<\/p>\n

3) The probability that someone in the U.S. is a Muslim, which is about 1%<\/a>.<\/p>\n

Putting it all together gives the following:<\/p>\n

Pr(Terrorist | Muslim) = [Pr(Muslim | Terrorist) x Pr(Terrorist)] \u00f7 Pr(Muslim) = 10% x 0.002% \u00f7 1% = 0.0002 or 0.02%.<\/p>\n

One interpretation of this result is that the probability that a Muslim person, whom you encounter at random in the U.S., is a terrorist is about 1\/50th of one-percent. In other words, around one in 5,000 Muslim persons you meet at random is a terrorist. \u00a0And keep in mind that the values I chose to make this calculation deliberately over-state, probably by a lot, that probability, so that the probability that a Muslim person is a Terrorist is likely much lower than 0.02%.<\/p>\n

Moreover, the probability that a Muslim person is a Terrorist (0.002%) is 500 times lower than the probability that a Terrorist is a Muslim (10%).<\/p>\n

(William Easterly of New York University applies Bayes\u2019 theorem to the same question<\/a>, using estimates that don\u2019t over-state as much as mine do, and calculates the difference not at 500 times but 13,000 times lower!)<\/p>\n

Other Considerations<\/strong><\/p>\n

As low as the probability of a Muslim person being a Terrorist is, the same data do indicate that a Non-Muslim person is much less likely to be a Terrorist. \u00a0By substituting values where appropriate \u2013 Pr(Non-Muslim | Terrorist) = 90% and Pr(Non-Muslim) = 99% \u2013 Bayes\u2019 theorem gives us the following:<\/p>\n

Pr(Terrorist | Non-Muslim) = [Pr(Non-Muslim | Terrorist) x Pr(Terrorist) \u00f7 Pr(Non-Muslim) = 90% x 0.002% \u00f7 99% = 0.00002 or 0.002%.<\/p>\n

So one interpretation of this is that a randomly chosen Non-Muslim person is around one-tenth as likely to be a Terrorist than a Muslim person (i.e. 0.2%\/0.002%). \u00a0Naturally, the probabilities will be higher or lower if you\u2019re at a terrorist convention or at an anti-terrorist peace rally; or if you have additional data that further differentiates among various groups \u2013 such as Wahhabi Sunni Muslims versus Salafist Muslim or Tamil Buddhists versus Tibetan Buddhists \u2013 the results again will be more accurate.<\/p>\n

But whether you\u2019re trying to educate yourself about the flu or terrorism, common sense suggests using relevant information as best you can. Bayes\u2019 theorem is a good way to do that.<\/p>\n

(I wish to thank Roger Koppl for helping me with an earlier version of this essay. Any remaining errors, however, are mine, alone.)<\/p>\n

Sanford (Sandy) Ikeda is a professor of economics at Purchase College, SUNY, and the author of The Dynamics of the Mixed Economy: Toward a Theory of Interventionism<\/em>.\u00a0He is a member of the FEE\u00a0Faculty Network<\/a>.<\/strong><\/p>\n

This article was published by The Foundation for Economic Education<\/a> and may be freely distributed, subject to a Creative Commons Attribution 4.0 International License<\/a>, which requires that credit be given to the author. Read the original article<\/a>.<\/em>
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Sanford Ikeda ****************************** It\u2019s flu season and for the past two days you\u2019ve had a headache and sore throat. You learn that 90% of people who actually have the flu also have those symptoms, which makes you worry. \u00a0Does that mean the chances of your having the flu is 90%? \u00a0In other words, if there\u2019s a 90% chance of having a headache and sore throat given that you have the flu, does that mean there\u2019s a 90% chance having the…<\/p>\n

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