The joint probability mass function of X and Y is given byp(1,1) = 0.35 p(1,2) = 0.05 p(1,3) = 0.05p(2,1) = 0.05 p(2,2) = 0Р(3, 1) — 0.1p(2,3) = 0.1p(3, 2) = 0.05 p(3,3) = 0.25(a) Compute the conditional mass function of Y given X = 2: P(Y = 1|X = 2)P(Y = 2|X = 2) =P(Y = 3|X = 2) =(b) Are X and Y independent? (enter YES or NO)

We will use the following formula for the conditional probability to solve the problems :

Answered: The joint probability mass function of…

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

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2. Let X and Y be two independent exponential random variables with parameters both equal to 3. (a) Find f(x, y), the joint probability density function of X and Y. (b) Let A = {(x, y) = R² : x > 4 and y > 5}. Find P(A). (c) Let B = {(x, y) = R² : x + y ≤ 2}. Find P(B).
The random variables X and Y have ä joint Probability masS 6 function as Follows - (X.Y); For ( X,Y) E A 18 For the other (x,Y) while A = {(x,Y ,2,3 end y=l,2} ore Depine: %3D a The probability mass Function For the random variable V= X+ Y b) The probability mass Function For the random variable V= X-Y 285
Q. 4 Let X and Y be two continuous random variables with following joint probability density function SK (x² + y²); 0 Y), (c) P(X +Y > 1) (b)
The joint probability mass function of Xand Ylis given by p(1, 1) = 0.2 p(2, 1) = 0.1 p(3, 1) = 0.05 p(3, 2) = 0.1 Р(1,2) - 0.1 p(2, 2) = 0.05 p(2,3) 0.05 p(1, 3) = 0.1 p(3, 3) = 0.25 %3D (a) Compute the conditional mass function of Ylgiven X 2: P(Y = 1|X = 2) D %3D P(Y = 2|X 2)= P(Y 3|X = 2) (b) Are Xland Yindependent? (enter YES or NO) NO
The joint probability mass function of X and Y is given by P(1, 1) = 0 p(2, 1) = 0.05 p(3, 1) = 0.1 (a) Compute the conditional mass function of Y given X = 2: P(Y = 1|X = 2) = P(Y = :2|X = 2) = P(Y = 3|X = 2) = (b) Are X and Y independent? (enter YES or NO) (c) Compute the following probabilities: P(X+Y> 2) = P(XY = 3) = P(> 1) = p(1,2)= 0.1 p(2, 2) = 0.3 p(3,2)= 0.05 p(1,3)= 0.05 p(2, 3) = 0.1 p(3, 3) = 0.25
X and Y are independent, X ~ U (0, 1) and Y U (0, 1). Calculate the probability P(X+Y<1.5).
Suppose that p(x, y), the joint probability mass function of X and Y, is given by p(1, 1) = 0.5, p(1,2)= 0.1, p(2, 1) = 0.1, p(2, 2) = 0.3 Calculate the conditional probability mass function of X given that Y = 1.
In a car service station, there are two service lines. The random variables X and Y are the proportions of time that line 1 and line 2 are in use, respectively. The joint probability density function for (X, Y) is given by S{(r² + y°), Osx s1;0sy s1 {6. f(x, y) = elsewhere. (a) If Z=X+ Y, the sum of the two proportions, find E(Z) ; (b) Find E(XY ); (c) Find Var(X); (d) Find Cov(X, Y );
The joint probability mass function of X and Y is given by p(1, 1) = 0.05 p(2, 1) = 0.1 p(1,2) = 0.05 p(1, 3) = 0.05 p(2,2) = 0 p(2, 3) = 0.05 p(3, 1) = 0.05 p(3,2) = 0.1 p(3, 3) = 0.55 (a) Compute the conditional mass function of Y given X = 2: P(Y = 1|X = 2) = ☐ P(Y = 2X = 2) = P(Y = 3|X = 2) (b) Are X and Y independent? (enter YES or NO) (c) Compute the following probabilities: P(X + Y > 2) = P(XY = 2) = P(¥ > 1) = |
Use a Venn diagram. Let P(Z) = 0.44, P(Y) = 0.28, and P(Z U Y) = 0.57. Find each probability. (a) P (Z'n Y') (b) P(Z'U Y) (c) P (Z' U Y) (d) P (Zn Y')

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