...one of the most highly
regarded and expertly designed C++ library projects in the
world.

— Herb Sutter and Andrei
Alexandrescu, C++
Coding Standards

A multiple choice test has four possible answers to each of 16 questions. A student guesses the answer to each question, so the probability of getting a correct answer on any given question is one in four, a quarter, 1/4, 25% or fraction 0.25. The conditions of the binomial experiment are assumed to be met: n = 16 questions constitute the trials; each question results in one of two possible outcomes (correct or incorrect); the probability of being correct is 0.25 and is constant if no knowledge about the subject is assumed; the questions are answered independently if the student's answer to a question in no way influences his/her answer to another question.

First, we need to be able to use the binomial distribution constructor (and some std input/output, of course).

#include <boost/math/distributions/binomial.hpp> using boost::math::binomial; #include <iostream> using std::cout; using std::endl; using std::ios; using std::flush; using std::left; using std::right; using std::fixed; #include <iomanip> using std::setw; using std::setprecision;

The number of correct answers, X, is distributed as a binomial random variable with binomial distribution parameters: questions n = 16 and success fraction probability p = 0.25. So we construct a binomial distribution:

int questions = 16; // All the questions in the quiz. int answers = 4; // Possible answers to each question. double success_fraction = (double)answers / (double)questions; // If a random guess. // Caution: = answers / questions would be zero (because they are integers)! binomial quiz(questions, success_fraction);

and display the distribution parameters we used thus:

cout << "In a quiz with " << quiz.trials() << " questions and with a probability of guessing right of " << quiz.success_fraction() * 100 << " %" << " or 1 in " << static_cast<int>(1. / quiz.success_fraction()) << endl;

Show a few probabilities of just guessing:

cout << "Probability of getting none right is " << pdf(quiz, 0) << endl; // 0.010023 cout << "Probability of getting exactly one right is " << pdf(quiz, 1) << endl; cout << "Probability of getting exactly two right is " << pdf(quiz, 2) << endl; int pass_score = 11; cout << "Probability of getting exactly " << pass_score << " answers right by chance is " << pdf(quiz, questions) << endl;

Probability of getting none right is 0.0100226 Probability of getting exactly one right is 0.0534538 Probability of getting exactly two right is 0.133635 Probability of getting exactly 11 answers right by chance is 2.32831e-010

These don't give any encouragement to guessers!

We can tabulate the 'getting exactly right' ( == ) probabilities thus:

cout << "\n" "Guessed Probability" << right << endl; for (int successes = 0; successes <= questions; successes++) { double probability = pdf(quiz, successes); cout << setw(2) << successes << " " << probability << endl; } cout << endl;

Guessed Probability 0 0.0100226 1 0.0534538 2 0.133635 3 0.207876 4 0.225199 5 0.180159 6 0.110097 7 0.0524273 8 0.0196602 9 0.00582526 10 0.00135923 11 0.000247132 12 3.43239e-005 13 3.5204e-006 14 2.51457e-007 15 1.11759e-008 16 2.32831e-010

Then we can add the probabilities of some 'exactly right' like this:

cout << "Probability of getting none or one right is " << pdf(quiz, 0) + pdf(quiz, 1) << endl;

Probability of getting none or one right is 0.0634764

But if more than a couple of scores are involved, it is more convenient (and may be more accurate) to use the Cumulative Distribution Function (cdf) instead:

cout << "Probability of getting none or one right is " << cdf(quiz, 1) << endl;

Probability of getting none or one right is 0.0634764

Since the cdf is inclusive, we can get the probability of getting up to 10 right ( <= )

cout << "Probability of getting <= 10 right (to fail) is " << cdf(quiz, 10) << endl;

Probability of getting <= 10 right (to fail) is 0.999715

To get the probability of getting 11 or more right (to pass), it is tempting to use

1 - cdf(quiz, 10)

to get the probability of > 10

cout << "Probability of getting > 10 right (to pass) is " << 1 - cdf(quiz, 10) << endl;

Probability of getting > 10 right (to pass) is 0.000285239

But this should be resisted in favor of using the complement function. Why complements?

cout << "Probability of getting > 10 right (to pass) is " << cdf(complement(quiz, 10)) << endl;

Probability of getting > 10 right (to pass) is 0.000285239

And we can check that these two, <= 10 and > 10, add up to unity.

BOOST_ASSERT((cdf(quiz, 10) + cdf(complement(quiz, 10))) == 1.);

If we want a < rather than a <= test, because the CDF is inclusive, we must subtract one from the score.

cout << "Probability of getting less than " << pass_score << " (< " << pass_score << ") answers right by guessing is " << cdf(quiz, pass_score -1) << endl;

Probability of getting less than 11 (< 11) answers right by guessing is 0.999715

and similarly to get a >= rather than a > test we also need
to subtract one from the score (and can again check the sum is unity).
This is because if the cdf is *inclusive*, then
its complement must be *exclusive* otherwise there
would be one possible outcome counted twice!

cout << "Probability of getting at least " << pass_score << "(>= " << pass_score << ") answers right by guessing is " << cdf(complement(quiz, pass_score-1)) << ", only 1 in " << 1/cdf(complement(quiz, pass_score-1)) << endl; BOOST_ASSERT((cdf(quiz, pass_score -1) + cdf(complement(quiz, pass_score-1))) == 1);

Probability of getting at least 11 (>= 11) answers right by guessing is 0.000285239, only 1 in 3505.83

Finally we can tabulate some probabilities:

cout << "\n" "At most (<=)""\n""Guessed OK Probability" << right << endl; for (int score = 0; score <= questions; score++) { cout << setw(2) << score << " " << setprecision(10) << cdf(quiz, score) << endl; } cout << endl;

At most (<=) Guessed OK Probability 0 0.01002259576 1 0.0634764398 2 0.1971110499 3 0.4049871101 4 0.6301861752 5 0.8103454274 6 0.9204427481 7 0.9728700437 8 0.9925302796 9 0.9983555346 10 0.9997147608 11 0.9999618928 12 0.9999962167 13 0.9999997371 14 0.9999999886 15 0.9999999998 16 1

cout << "\n" "At least (>)""\n""Guessed OK Probability" << right << endl; for (int score = 0; score <= questions; score++) { cout << setw(2) << score << " " << setprecision(10) << cdf(complement(quiz, score)) << endl; }

At least (>) Guessed OK Probability 0 0.9899774042 1 0.9365235602 2 0.8028889501 3 0.5950128899 4 0.3698138248 5 0.1896545726 6 0.07955725188 7 0.02712995629 8 0.00746972044 9 0.001644465374 10 0.0002852391917 11 3.810715862e-005 12 3.783265129e-006 13 2.628657967e-007 14 1.140870154e-008 15 2.328306437e-010 16 0

We now consider the probabilities of **ranges**
of correct guesses.

First, calculate the probability of getting a range of guesses right, by adding the exact probabilities of each from low ... high.

int low = 3; // Getting at least 3 right. int high = 5; // Getting as most 5 right. double sum = 0.; for (int i = low; i <= high; i++) { sum += pdf(quiz, i); } cout.precision(4); cout << "Probability of getting between " << low << " and " << high << " answers right by guessing is " << sum << endl; // 0.61323

Probability of getting between 3 and 5 answers right by guessing is 0.6132

Or, usually better, we can use the difference of cdfs instead:

cout << "Probability of getting between " << low << " and " << high << " answers right by guessing is " << cdf(quiz, high) - cdf(quiz, low - 1) << endl; // 0.61323

Probability of getting between 3 and 5 answers right by guessing is 0.6132

And we can also try a few more combinations of high and low choices:

low = 1; high = 6; cout << "Probability of getting between " << low << " and " << high << " answers right by guessing is " << cdf(quiz, high) - cdf(quiz, low - 1) << endl; // 1 and 6 P= 0.91042 low = 1; high = 8; cout << "Probability of getting between " << low << " and " << high << " answers right by guessing is " << cdf(quiz, high) - cdf(quiz, low - 1) << endl; // 1 <= x 8 P = 0.9825 low = 4; high = 4; cout << "Probability of getting between " << low << " and " << high << " answers right by guessing is " << cdf(quiz, high) - cdf(quiz, low - 1) << endl; // 4 <= x 4 P = 0.22520

Probability of getting between 1 and 6 answers right by guessing is 0.9104 Probability of getting between 1 and 8 answers right by guessing is 0.9825 Probability of getting between 4 and 4 answers right by guessing is 0.2252

Using moments of the distribution, we can say more about the spread of results from guessing.

cout << "By guessing, on average, one can expect to get " << mean(quiz) << " correct answers." << endl; cout << "Standard deviation is " << standard_deviation(quiz) << endl; cout << "So about 2/3 will lie within 1 standard deviation and get between " << ceil(mean(quiz) - standard_deviation(quiz)) << " and " << floor(mean(quiz) + standard_deviation(quiz)) << " correct." << endl; cout << "Mode (the most frequent) is " << mode(quiz) << endl; cout << "Skewness is " << skewness(quiz) << endl;

By guessing, on average, one can expect to get 4 correct answers. Standard deviation is 1.732 So about 2/3 will lie within 1 standard deviation and get between 3 and 5 correct. Mode (the most frequent) is 4 Skewness is 0.2887

The quantiles (percentiles or percentage points) for a few probability levels:

cout << "Quartiles " << quantile(quiz, 0.25) << " to " << quantile(complement(quiz, 0.25)) << endl; // Quartiles cout << "1 standard deviation " << quantile(quiz, 0.33) << " to " << quantile(quiz, 0.67) << endl; // 1 sd cout << "Deciles " << quantile(quiz, 0.1) << " to " << quantile(complement(quiz, 0.1))<< endl; // Deciles cout << "5 to 95% " << quantile(quiz, 0.05) << " to " << quantile(complement(quiz, 0.05))<< endl; // 5 to 95% cout << "2.5 to 97.5% " << quantile(quiz, 0.025) << " to " << quantile(complement(quiz, 0.025)) << endl; // 2.5 to 97.5% cout << "2 to 98% " << quantile(quiz, 0.02) << " to " << quantile(complement(quiz, 0.02)) << endl; // 2 to 98% cout << "If guessing then percentiles 1 to 99% will get " << quantile(quiz, 0.01) << " to " << quantile(complement(quiz, 0.01)) << " right." << endl;

Notice that these output integral values because the default policy
is `integer_round_outwards`

.

Quartiles 2 to 5 1 standard deviation 2 to 5 Deciles 1 to 6 5 to 95% 0 to 7 2.5 to 97.5% 0 to 8 2 to 98% 0 to 8

Quantiles values are controlled by the discrete
quantile policy chosen. The default is `integer_round_outwards`

,
so the lower quantile is rounded down, and the upper quantile is
rounded up.

But we might believe that the real values tell us a little more - see Understanding Discrete Quantile Policy.

We could control the policy for **all**
distributions by

#define BOOST_MATH_DISCRETE_QUANTILE_POLICY real at the head of the program would make this policy apply

#define BOOST_MATH_DISCRETE_QUANTILE_POLICY real

at the head of the program would make this policy apply

at the head of the program would make this policy apply to this
**one, and only**, translation unit.

Or we can now create a (typedef for) policy that has discrete quantiles real (here avoiding any 'using namespaces ...' statements):

using boost::math::policies::policy; using boost::math::policies::discrete_quantile; using boost::math::policies::real; using boost::math::policies::integer_round_outwards; // Default. typedef boost::math::policies::policy<discrete_quantile<real> > real_quantile_policy;

Add a custom binomial distribution called

`real_quantile_binomial`

that uses

`real_quantile_policy`

using boost::math::binomial_distribution; typedef binomial_distribution<double, real_quantile_policy> real_quantile_binomial;

Construct an object of this custom distribution:

real_quantile_binomial quiz_real(questions, success_fraction);

And use this to show some quantiles - that now have real rather than integer values.

cout << "Quartiles " << quantile(quiz, 0.25) << " to " << quantile(complement(quiz_real, 0.25)) << endl; // Quartiles 2 to 4.6212 cout << "1 standard deviation " << quantile(quiz_real, 0.33) << " to " << quantile(quiz_real, 0.67) << endl; // 1 sd 2.6654 4.194 cout << "Deciles " << quantile(quiz_real, 0.1) << " to " << quantile(complement(quiz_real, 0.1))<< endl; // Deciles 1.3487 5.7583 cout << "5 to 95% " << quantile(quiz_real, 0.05) << " to " << quantile(complement(quiz_real, 0.05))<< endl; // 5 to 95% 0.83739 6.4559 cout << "2.5 to 97.5% " << quantile(quiz_real, 0.025) << " to " << quantile(complement(quiz_real, 0.025)) << endl; // 2.5 to 97.5% 0.42806 7.0688 cout << "2 to 98% " << quantile(quiz_real, 0.02) << " to " << quantile(complement(quiz_real, 0.02)) << endl; // 2 to 98% 0.31311 7.7880 cout << "If guessing, then percentiles 1 to 99% will get " << quantile(quiz_real, 0.01) << " to " << quantile(complement(quiz_real, 0.01)) << " right." << endl;

Real Quantiles Quartiles 2 to 4.621 1 standard deviation 2.665 to 4.194 Deciles 1.349 to 5.758 5 to 95% 0.8374 to 6.456 2.5 to 97.5% 0.4281 to 7.069 2 to 98% 0.3131 to 7.252 If guessing then percentiles 1 to 99% will get 0 to 7.788 right.

See binomial_quiz_example.cpp for full source code and output.