Continuous event probability table

Continuous event probability chart with two different colors balls, cards or different objects

% f_tabpro - all convinations possible to draw consecutively from the box
% with x balls ,number card or other object and y numbers of the
% complemente object
% (z) times . The box have x balls and y balls with 2 diferents color.
% [ omp,omm ] = f_tabpro( x, y, z) where x and y are non-negative integers
% and number of total balls of diferentes colors
% z is the number times of we draw balls from the box consecutively .
% omp is probability table for consecutive acctions
% omn is the convinations table of darw balls ,
% 1 = firts color or ases in deck of cards
% 0 = the other color or the rest of the deck of cards

%
% Examples:
% [ omp,omn] = f_tabpro( 2,3,3)
% omp =
% 0.6000 0.5000 0.3333
% 0.4000 0.7500 0.6667
% 0.6000 0.5000 0.6667
% 0.6000 0.5000 0.6667
% 0.4000 0.2500 1.0000
% 0.4000 0.7500 0.3333
% 0.6000 0.5000 0.3333
%
% omn =
% 0 0 0
% 1 0 0
% 0 1 0
% 0 0 1
% 1 1 0
% 1 0 1
% 0 1 1

% usefull sample in deck of cards sample
%%
% In a deck of 20 cards( 1,1,2,2,3,3,4,4........9,9,10,10) we choose 3 cards and have now a deck whit 15 cards,
% What is the probability of getting an ace ("1") in the first attempt in this 15 remaining cards

% total =20 cards....
% x= 2 ACE
% y= 18 no ACE , the rest
% total darw card 3 + last one = 4
% z=4;

%[ omp,omm ] = f_tabpro( 2,18,4)
% omp =
%
% 0.9000 0.8947 0.8889 0.8824
% 0.1000 0.9474 0.9444 0.9412
% 0.9000 0.1053 0.9444 0.9412
% 0.9000 0.8947 0.1111 0.9412
% 0.9000 0.8947 0.8889 0.1176
% 0.1000 0.0526 1.0000 1.0000
% 0.1000 0.9474 0.0556 1.0000
% 0.1000 0.9474 0.9444 0.0588
% 0.9000 0.1053 0.0556 1.0000
% 0.9000 0.1053 0.9444 0.0588
% 0.9000 0.8947 0.1111 0.0588

% omm =
%
% 0 0 0 0->
% 1 0 0 0->
% 0 1 0 0->
% 0 0 1 0->
% 0 0 0 1->
% 1 1 0 0->
% 1 0 1 0->
% 1 0 0 1->
% 0 1 1 0->
% 0 1 0 1->
% 0 0 1 1->

% file 1.... 0 0 0 0 = 0.9000 * 0.8947 * 0.8889 * 0.8824 = 06316
% file 2......1 0 0 0 = 0.1000 * 0.9474 * 0.9444 * 0.9412 =0.0842
% file.....
% ..
% ..
% file 11... 0 0 1 1 = 0.9000 * 0.8947 * 0.1111 * 0.0588= 0.0053
%
% SUM res=06316+0.0842+...... ......+0.0053 = 1
% What is the probability of getting an ace ("1") in the first attempt in this 15
% remaining cards = SUM of Prod File prob where last number is 1

% See the table
% Last4 =0 Last4= 1,
% ----------------------
% 0 0 0 0 Prob-> 0.9000 * 0.8947 * 0.8889 * 0.8824 = 0.6316 0.6316
% 1 0 0 0 Prob-> 0.1000 * 0.9474 * 0.9444 * 0.9412 = 0.0842 0.0842
% 0 1 0 0 Prob-> 0.9000 * 0.1053 * 0.9444 * 0.9412 = 0.0842 0.0842
% 0 0 1 0 Prob-> 0.9000 * 0.8947 * 0.1111 * 0.9412 = 0.0842 0.0842
% 0 0 0 1 Prob-> 0.9000 * 0.8947 * 0.8889 * 0.1176 = 0.0842 - 0.0842
% 1 0 1 0 Prob-> 0.1000 * 0.0526 * 1.0000 * 1.0000 = 0.0053 0.0053
% 1 0 1 0 Prob-> 0.1000 * 0.9474 * 0.0556 * 1.0000 = 0.0053 0.0053
% 1 0 0 1 Prob-> 0.1000 * 0.9474 * 0.9444 * 0.0588 = 0.0053 - 0.0053
% 0 1 1 0 Prob-> 0.9000 * 0.1053 * 0.0556 * 1.0000 = 0.0053 0.0053
% 0 1 0 1 Prob-> 0.9000 * 0.1053 * 0.9444 * 0.0588 = 0.0053 - 0.0053
% 0 0 1 1 Prob-> 0.9000 * 0.8947 * 0.1111 * 0.0588 = 0.0053 - 0.0053
% ---------------------------------------------------------------------------------------------
% ----------------------------------------------- SUM>= 1.0000 0.9000 0.1000

%%
% use the function
% f_true_table( x,y,z); By Aldo Tamariz [e-mail]

% for Matlab R2016a

% (c) Aldo Tamariz B
% email: [e-mail]