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Two cards are selected at random, without replacement, from a standard deck. Let X be the number of aces and Y be the number of spades. Under the usual assumptions, determine the joint distribution and the marginals.

Let X be the number of aces and Y be the number of spades. Define the events A S i , A i , S i , and N i , i = 1 , 2 , of drawing ace of spades, other ace, spade (other than the ace), and neither on the i selection. Let P ( i , k ) = P ( X = i , Y = k ) .

P ( 0 , 0 ) = P ( N 1 N 2 ) = 36 52 35 51 = 1260 2652

P ( 0 , 1 ) = P ( N 1 S 2 S 1 N 2 ) = 36 52 12 51 + 12 52 36 51 = 864 2652

P ( 0 , 2 ) = P ( S 1 S 2 ) = 12 52 11 51 = 132 2652

P ( 1 , 0 ) = P ( A 1 N 2 N 1 S 2 ) = 3 52 36 51 + 36 52 3 51 = 216 2652

P ( 1 , 1 ) = P ( A 1 S 2 S 1 A 2 A S 1 N 2 N 1 A S 2 ) = 3 52 12 51 + 12 52 3 51 + 1 52 36 51 + 36 52 1 51 = 144 2652

P ( 1 , 2 ) = P ( A S 1 S 2 S 1 A S 2 ) = 1 52 12 51 + 12 52 1 51 = 24 2652

P ( 2 , 0 ) = P ( A 1 A 2 ) = 3 52 2 51 = 6 2652

P ( 2 , 1 ) = P ( A S 1 A 2 A 1 A S 2 ) = 1 52 3 51 + 3 52 1 51 = 6 2652

P ( 2 , 2 ) = P ( ) = 0

% type npr08_01 % file npr08_01.m % Solution for [link] X = 0:2; Y = 0:2;Pn = [132 24 0; 864 144 6; 1260 216 6];P = Pn/(52*51); disp('Data in Pn, P, X, Y')npr08_01 % Call for mfileData in Pn, P, X, Y % Result PX = sum(P)PX = 0.8507 0.1448 0.0045 PY = fliplr(sum(P'))PY = 0.5588 0.3824 0.0588
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Two positions for campus jobs are open. Two sophomores, three juniors, and three seniors apply. It is decided to select two at random (each possible pairequally likely). Let X be the number of sophomores and Y be the number of juniors who are selected. Determine the joint distribution for the pair { X , Y } and from this determine the marginals for each.

Let A i , B i , C i be the events of selecting a sophomore, junior, or senior, respectively, on the i th trial. Let X be the number of sophomores and Y be the number of juniors selected.

Set P ( i , k ) = P ( X = i , Y = k )

P ( 0 , 0 ) = P ( C 1 C 2 ) = 3 8 2 7 = 6 56

P ( 0 , 1 ) = P ( B 1 C 2 ) + P ( C 1 B 2 ) = 3 8 3 7 + 3 8 3 7 = 18 56

P ( 0 , 2 ) = P ( B 1 B 2 ) = 3 8 2 7 = 6 56

P ( 1 , 0 ) = P ( A 1 C 2 ) + P ( C 1 A 2 ) = 2 8 3 7 + 3 8 2 7 = 12 56

P ( 1 , 1 ) = P ( A 1 B 2 ) + P ( B 1 A 2 ) = 2 8 3 7 + 3 8 2 7 = 12 56

P ( 2 , 0 ) = P ( A 1 A 2 ) = 2 8 1 7 = 2 56

P ( 1 , 2 ) = P ( 2 , 1 ) = P ( 2 , 2 ) = 0

P X = [ 30 / 56 24 / 56 2 / 56 ] P Y = [ 20 / 56 30 / 56 6 / 56 ]

% file npr08_02.m % Solution for [link] X = 0:2; Y = 0:2;Pn = [6 0 0; 18 12 0; 6 12 2];P = Pn/56; disp('Data are in X, Y,Pn, P') npr08_02 Data are in X, Y,Pn, P PX = sum(P)PX = 0.5357 0.4286 0.0357 PY = fliplr(sum(P'))PY = 0.3571 0.5357 0.1071
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A die is rolled. Let X be the number that turns up. A coin is flipped X times. Let Y be the number of heads that turn up. Determine the joint distribution for the pair { X , Y } . Assume P ( X = k ) = 1 / 6 for 1 k 6 and for each k , P ( Y = j | X = k ) has the binomial ( k , 1 / 2 ) distribution. Arrange the joint matrix as on the plane, with values of Y increasing upward. Determine the marginal distribution for Y . (For a MATLAB based way to determine the joint distribution see Example 7 from "Conditional Expectation, Regression")

P ( X = i , Y = k ) = P ( X = i ) P ( Y = k | X = i ) = ( 1 / 6 ) P ( Y = k | X = i ) .

% file npr08_03.m % Solution for [link] X = 1:6; Y = 0:6;P0 = zeros(6,7); % Initialize for i = 1:6 % Calculate rows of Y probabilitiesP0(i,1:i+1) = (1/6)*ibinom(i,1/2,0:i); endP = rot90(P0); % Rotate to orient as on the plane PY = fliplr(sum(P')); % Reverse to put in normal orderdisp('Answers are in X, Y, P, PY') npr08_03 % Call for solution m-file Answers are in X, Y, P, PYdisp(P) 0 0 0 0 0 0.00260 0 0 0 0.0052 0.0156 0 0 0 0.0104 0.0260 0.03910 0 0.0208 0.0417 0.0521 0.0521 0 0.0417 0.0625 0.0625 0.0521 0.03910.0833 0.0833 0.0625 0.0417 0.0260 0.0156 0.0833 0.0417 0.0208 0.0104 0.0052 0.0026disp(PY) 0.1641 0.3125 0.2578 0.1667 0.0755 0.0208 0.0026
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Source:  OpenStax, Applied probability. OpenStax CNX. Aug 31, 2009 Download for free at http://cnx.org/content/col10708/1.6
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