Probability of Success with Random GuessingDate: 07/12/2008 at 12:44:54 From: Tom Subject: Probability: Why 75% I'm reading a book called "The Drunkard's Walk" by Dr. Leonard Mlodinow. It's an entirely readable look at the history of probability and the misconceptions of probability. I have come to a part that I don't understand: "In solving the problem, Galileo implicitly employed our next important principle: The chances of an event depend on the number of ways in which it can occur. That is not a surprising statement. The surprise is just how large that effect is--and how difficult it can be to calculate. For example, suppose you give a 10-question true-or-false quiz to your class of 25 sixth-graders. Let's do an accounting of the results a particular student might achieve: she could answer all questions correctly; she could miss 1 question--that can happen in 10 ways because there are 10 questions she could miss; she could miss a pair of questions--that can happen in 45 ways because there are 45 distinct pairs of questions; and so on. As a result, on average in a collection of students who are randomly guessing, for every student scoring 100 percent, you'll find about 10 scoring 90 percent and 45 scoring 80 percent. The chances of getting a grade near 50 percent are of course still higher still, but in a class of 25 the probability that at least one student will get a B (80 percent) or better if all the students are guessing is about 75 percent. So, if you are a veteran teacher, it is likely that among all the students over the years who have shown up unprepared and more or less guessed at your quizzes, some were rewarded with an A or B." I believe I understand how he calculated his other numbers, but how does he come up with the 75%? That seems too high to me. Date: 07/14/2008 at 14:13:29 From: Doctor Achilles Subject: Re: Probability: Why 75% Hi Tom, Thanks for writing to Dr. Math. The total number of possible ways to answer 10 true/false questions is 2^10 or 1024. If students are randomly guessing, the number of ways to get 100% is 1. The number of ways to get 90% is 10. The number of ways to get 80% is 45. Therefore, the total number of ways that a randomly guessing student can get an 80% or better is 56. So the probability of any given student getting 80% or better when guessing randomly is 56/1024 or 7/128 (or a little better than 5%). If we turn that around, the chances that a single student will get worse than 80% is 121/128. Now, if we have 2 students who are randomly guessing, what is the chance that BOTH of them will get worse than 80%? That's (121/128)^2 or 14641/16384 (or about 89%). If we have 3 students randomly guessing, the probability that all three get worse than 80% is (121/128)^3 = about 84%. With 4 students randomly guessing, the probability of all four get worse than 80% (in other words, the probability that NONE of them will get 80% or better is (121/128)^4 = 80%. So, in other words, with 4 students guessing randomly, there is a 20% chance that at least one will walk away with a B or better. With 5 guessing, the probability that none get a B or better is (121/128)^5 = 75%, so there is a 25% chance that one walks away with a B or better. With n students guessing, the probability of none of them getting a B or better is (121/128)^n = P. Let's ask how many students we need to get a P value of 25%. For that, we have: (121/128)^n = 0.25 Let's take the log of both sides: log ((121/128)^n) = log (0.25) Using the properties of logarithms, What is Log? http://mathforum.org/library/drmath/view/58046.html Logarithms: Solving for T http://mathforum.org/library/drmath/view/55561.html we can convert log ((121/128)^n) = log (0.25) to n * log (121/128) = log (0.25) , so n = log (0.25) / (log (121/128)) . Using my calculator I get: n = 24.6 So we can go back and check. With 24 students guessing, the probability that NONE will get a B or better is (121/128)^24 = 25.9%. So there is a 74.1% probability that one or more of the 24 guessers walks away with a B or better. With 25 students guessing, the probability that NONE will get a B or better is (121/128)^25 = 24.5%. So the probability that AT LEAST one student walks away with a B or better is 75.5%. So that means if you give your test to 200 students, and 25 of those 200 students never studied at all and just guess randomly, there is more than a 75% chance that at least one of those random guessers walks away with a B or better. I went pretty quickly over a lot of this, so if you want to talk about any of the logic steps in more detail, please write back and let me know. - Doctor Achilles, The Math Forum http://mathforum.org/dr.math/ Date: 07/14/2008 at 15:39:07 From: Tom Subject: Thank you (Probability: Why 75%) Thank you so much for your great explanation! I had to read it several times before I "got it" but I now understand. Thanks! |
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