Guessing on True-False TestsDate: 07/08/99 at 18:13:25 From: Annette Darby Subject: Using binomial theorem to calculate probability An examination is comprised of n true-false questions. To pass, you must answer at least 1/2 correctly. Suppose you guess at each answer. Are you more likely to pass a test with a few questions than you are to pass a test with many questions? Does it matter if the number of questions is odd or even? I know the answers are: few questions, and yes, it does matter, but I don't know why. Thank you. Date: 07/08/99 at 19:41:03 From: Doctor Anthony Subject: Re: Using binomial theorem to calculate probability Suppose we have 8 questions. The probability of at least 4 correct is: P(4) = C(8,4) x (1/2)^4 x (1/2)^4 = 70 x (1/2)^8 P(5) = C(8,5) x (1/2)^8 = 56 x (1/2)^8 P(6) = C(8,6) x (1/2)^8 = 28 x (1/2)^8 P(7) = C(8,7) x (1/2)^8 = 8 x (1/2)^8 P(8) = C(8,8) x (1/2)^8 = 1 x (1/2)^8 ------------------- Total = 163 x (1/2)^8 = 0.63672 Suppose we have 6 questions. The probability of at least 3 correct is: P(3) = C(6,3) x (1/2)^6 = 20 x (1/2)^6 P(4) = C(6,4) x (1/2)^6 = 15 x (1/2)^6 P(5) = C(6,5) x (1/2)^6 = 6 x (1/2)^6 P(6) = C(6,6) x (1/2)^6 = 1 x (1/2)^6 ------------------- Total = 42 x (1/2)^6 = 0.65625 and with a smaller EVEN number of questions we have a better probability of passing. This is because the sum of the binomial coefficients with more questions does not compensate for the extra factors of (1/2) for the additional questions. Here we are comparing EVEN numbers of questions. With an odd number of questions the probability of passing is 1/2 whether you have 101 questions or only 1 question. If there is an odd number of questions, say 7, then you must get 4, 5, 6, 7 correct, which will work out as total probability = 1/2 P(4) = C(7,4) x (1/2)^7 = 35 x (1/2)^7 P(5) = C(7,5) x (1/2)^7 = 21 x (1/2)^7 P(6) = C(7,6) x (1/2)^7 = 7 x (1/2)^7 P(7) = C(7,7) x (1/2)^7 = 1 x (1/2)^7 ------------------- Total = 64 x (1/2)^7 = 0.5 You will find that for odd numbers of questions your probability of passing will NEVER vary from 1/2. This is because of the symmetry of the binomial coefficients in the expansion of (q+p)^n when n is odd. The coefficients when n = 3 are [C(3,0) + C(3,1) + C(3,2) + C(3,3)](1/2)^3 [ 1 + 3 + 3 + 1 ](1/2)^3 - Doctor Anthony, The Math Forum http://mathforum.org/dr.math/ |
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