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Guessing on True-False Tests


Date: 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/   
    
Associated Topics:
High School Probability

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