8.2 Problems on random vectors and joint distributions

Applied probability Page 1 / 6

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 $P (i, k) = P (X = i, Y = k)$ .

$P (0, 0) = P (N_{1} N_{2}) = \frac{36}{52} • \frac{35}{51} = \frac{1260}{2652}$

$P (0, 1) = P (N_{1} S_{2} ⋁ S_{1} N_{2}) = \frac{36}{52} • \frac{12}{51} + \frac{12}{52} • \frac{36}{51} = \frac{864}{2652}$

$P (0, 2) = P (S_{1} S_{2}) = \frac{12}{52} • \frac{11}{51} = \frac{132}{2652}$

$P (1, 0) = P (A_{1} N_{2} ⋁ N_{1} S_{2}) = \frac{3}{52} • \frac{36}{51} + \frac{36}{52} • \frac{3}{51} = \frac{216}{2652}$

$P (1, 1) = P (A_{1} S_{2} ⋁ S_{1} A_{2} ⋁ A S_{1} N_{2} ⋁ N_{1} A S_{2}) = \frac{3}{52} • \frac{12}{51} + \frac{12}{52} • \frac{3}{51} + \frac{1}{52} • \frac{36}{51} + \frac{36}{52} • \frac{1}{51} = \frac{144}{2652}$

$P (1, 2) = P (A S_{1} S_{2} ⋁ S_{1} A S_{2}) = \frac{1}{52} • \frac{12}{51} + \frac{12}{52} • \frac{1}{51} = \frac{24}{2652}$

$P (2, 0) = P (A_{1} A_{2}) = \frac{3}{52} • \frac{2}{51} = \frac{6}{2652}$

$P (2, 1) = P (A S_{1} A_{2} ⋁ A_{1} A S_{2}) = \frac{1}{52} • \frac{3}{51} + \frac{3}{52} • \frac{1}{51} = \frac{6}{2652}$

$P (2, 2) = P (\emptyset) = 0$

% 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 $P (i, k) = P (X = i, Y = k)$

$P (0, 0) = P (C_{1} C_{2}) = \frac{3}{8} • \frac{2}{7} = \frac{6}{56}$

$P (0, 1) = P (B_{1} C_{2}) + P (C_{1} B_{2}) = \frac{3}{8} • \frac{3}{7} + \frac{3}{8} • \frac{3}{7} = \frac{18}{56}$

$P (0, 2) = P (B_{1} B_{2}) = \frac{3}{8} • \frac{2}{7} = \frac{6}{56}$

$P (1, 0) = P (A_{1} C_{2}) + P (C_{1} A_{2}) = \frac{2}{8} • \frac{3}{7} + \frac{3}{8} • \frac{2}{7} = \frac{12}{56}$

$P (1, 1) = P (A_{1} B_{2}) + P (B_{1} A_{2}) = \frac{2}{8} • \frac{3}{7} + \frac{3}{8} • \frac{2}{7} = \frac{12}{56}$

$P (2, 0) = P (A_{1} A_{2}) = \frac{2}{8} • \frac{1}{7} = \frac{2}{56}$

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

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

% 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 $1 \leq k \leq 6$ 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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