Joint Probability Distribution of Two Events in a Random Experiment, Quizzes of Probability and Statistics

The solution to quiz 7 in math 5010-2, where the goal is to find the joint probability mass function, marginal probability mass functions, and determine the independence of two events related to the selection of balls from an urn. A table showing the joint distribution and the marginal distributions.

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Pre 2010

Uploaded on 08/30/2009

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Math 5010-2 Name:
Quiz 7
10/29/08
1. Six balls, labeled A, B, C, D, E, F, are in an urn. Two are selected
successively at random with the first one replaced before the second is drawn.
Let Xbe the number of times ball A is chosen and Ythe number of times ball
B or C is chosen. Obviously, 0 X2 and 0 Y2.
(a) Give the joint probability mass function of Xand Yin the form of a
table, with rows corresponding to xvalues (increasing from top to bottom) and
columns corresponding to yvalues (increasing from left to right). If correct, the
entries will sum to 1.
Table 1: Joint distribution.
x/y 0 1 2 totals
0 (3/6)22(2/6)(3/6) (2/6)2
1 2(1/6)(3/6) 2(1/6)(2/6) 0
2 (1/6)20 0
totals
(b) Find the marginal probability mass functions of Xand of Y, and be sure
to indicate which is which.
Table 2: Joint distribution with marginals. Marginal of Xis in right column,
that of Yis in bottom row.
x/y 0 1 2 totals
0 9/36 12/36 4/36 25/36
1 6/36 4/36 0 10/36
2 1/36 0 0 1/36
totals 16/36 16/36 4/36
(c) Are Xand Yindependent? Justify your answer.
No. For example 0 = P(X= 2, Y = 2) 6=P(X= 2)P(Y= 2) = 4/(36)2.
1

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Math 5010-2 Name: Quiz 7 10/29/

  1. Six balls, labeled A, B, C, D, E, F, are in an urn. Two are selected successively at random with the first one replaced before the second is drawn. Let X be the number of times ball A is chosen and Y the number of times ball B or C is chosen. Obviously, 0 ≤ X ≤ 2 and 0 ≤ Y ≤ 2.

(a) Give the joint probability mass function of X and Y in the form of a table, with rows corresponding to x values (increasing from top to bottom) and columns corresponding to y values (increasing from left to right). If correct, the entries will sum to 1.

Table 1: Joint distribution.

x/y 0 1 2 totals 0 (3/6)^2 2(2/6)(3/6) (2/6)^2 1 2(1/6)(3/6) 2(1/6)(2/6) 0 2 (1/6)^2 0 totals

(b) Find the marginal probability mass functions of X and of Y , and be sure to indicate which is which.

Table 2: Joint distribution with marginals. Marginal of X is in right column, that of Y is in bottom row.

x/y 0 1 2 totals

0 9 / 36 12 / 36 4 / 36 25/ 1 6 / 36 4 / 36 0 10/ 2 1 / 36 0 0 1/ totals 16/36 16/36 4/

(c) Are X and Y independent? Justify your answer. No. For example 0 = P (X = 2, Y = 2) 6 = P (X = 2)P (Y = 2) = 4/(36)^2.