Math Problem Statement

The probability mass function of the discrete random variables is given in the following table.

x y 0 1 2 0 0.05 0.15 0.075 1 0.045 0.05 0.025 2 0.1 0.04 0.065 3 0.2 0.05 0.15

Find P(X=0,Y=2) Marginal probability mass function of X. P(X>Y)

Solution

We are given the probability mass function (PMF) of two discrete random variables XX and YY in a tabular form. Let’s calculate the required probabilities step by step.

1. Find P(X=0,Y=2)P(X = 0, Y = 2):

From the table, the probability P(X=0,Y=2)P(X = 0, Y = 2) corresponds to the entry in the row X=0X = 0 and column Y=2Y = 2. From the table:

P(X=0,Y=2)=0.075P(X = 0, Y = 2) = 0.075


2. Find the marginal PMF of XX:

The marginal PMF of XX, denoted as P(X=x)P(X = x), is obtained by summing over all values of YY for each fixed X=xX = x:

P(X=x)=yP(X=x,Y=y)P(X = x) = \sum_{y} P(X = x, Y = y)

Calculation:

  • For X=0X = 0: P(X=0)=P(X=0,Y=0)+P(X=0,Y=1)+P(X=0,Y=2)=0.05+0.15+0.075=0.275P(X = 0) = P(X = 0, Y = 0) + P(X = 0, Y = 1) + P(X = 0, Y = 2) = 0.05 + 0.15 + 0.075 = 0.275

  • For X=1X = 1: P(X=1)=P(X=1,Y=0)+P(X=1,Y=1)+P(X=1,Y=2)=0.045+0.05+0.025=0.12P(X = 1) = P(X = 1, Y = 0) + P(X = 1, Y = 1) + P(X = 1, Y = 2) = 0.045 + 0.05 + 0.025 = 0.12

  • For X=2X = 2: P(X=2)=P(X=2,Y=0)+P(X=2,Y=1)+P(X=2,Y=2)=0.1+0.04+0.065=0.205P(X = 2) = P(X = 2, Y = 0) + P(X = 2, Y = 1) + P(X = 2, Y = 2) = 0.1 + 0.04 + 0.065 = 0.205

  • For X=3X = 3: P(X=3)=P(X=3,Y=0)+P(X=3,Y=1)+P(X=3,Y=2)=0.2+0.05+0.15=0.4P(X = 3) = P(X = 3, Y = 0) + P(X = 3, Y = 1) + P(X = 3, Y = 2) = 0.2 + 0.05 + 0.15 = 0.4

Thus, the marginal PMF of XX is: P(X=0)=0.275,P(X=1)=0.12,P(X=2)=0.205,P(X=3)=0.4P(X = 0) = 0.275, \quad P(X = 1) = 0.12, \quad P(X = 2) = 0.205, \quad P(X = 3) = 0.4


3. Find P(X>Y)P(X > Y):

To compute P(X>Y)P(X > Y), we sum the probabilities where X>YX > Y. This means summing over all table entries where the value of XX is strictly greater than the corresponding value of YY. From the table:

When X=1X = 1:

  • P(X=1,Y=0)=0.045P(X = 1, Y = 0) = 0.045

When X=2X = 2:

  • P(X=2,Y=0)=0.1P(X = 2, Y = 0) = 0.1
  • P(X=2,Y=1)=0.04P(X = 2, Y = 1) = 0.04

When X=3X = 3:

  • P(X=3,Y=0)=0.2P(X = 3, Y = 0) = 0.2
  • P(X=3,Y=1)=0.05P(X = 3, Y = 1) = 0.05
  • P(X=3,Y=2)=0.15P(X = 3, Y = 2) = 0.15

Now, sum these probabilities: P(X>Y)=0.045+0.1+0.04+0.2+0.05+0.15=0.585P(X > Y) = 0.045 + 0.1 + 0.04 + 0.2 + 0.05 + 0.15 = 0.585


Final Results:

  1. P(X=0,Y=2)=0.075P(X = 0, Y = 2) = 0.075
  2. Marginal PMF of XX: P(X=0)=0.275,P(X=1)=0.12,P(X=2)=0.205,P(X=3)=0.4P(X = 0) = 0.275, \quad P(X = 1) = 0.12, \quad P(X = 2) = 0.205, \quad P(X = 3) = 0.4
  3. P(X>Y)=0.585P(X > Y) = 0.585

Do you want me to verify the calculations or explain a specific part in more detail?


Related Questions:

  1. How is the joint PMF of XX and YY used to calculate marginal probabilities?
  2. What does P(X>Y)P(X > Y) represent in terms of real-world probability?
  3. How do we calculate P(Y>X)P(Y > X), and would it complement P(X>Y)P(X > Y)?
  4. Can we use marginal PMFs to check if XX and YY are independent?
  5. What if we were asked for P(XY)P(X \leq Y)? How would we compute it?

Tip:

When working with probabilities, always check if the total probability of all events sums to 1 to verify the correctness of your PMF table.

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Math Problem Analysis

Mathematical Concepts

Probability Theory
Discrete Random Variables
Joint Probability Mass Function
Marginal Probability Mass Function

Formulas

P(X = x, Y = y) for joint probability
P(X = x) = ∑ P(X = x, Y = y) for marginal probability
P(X > Y) = sum of probabilities where X > Y

Theorems

Law of Total Probability (for marginal PMF)
Basic Probability Rules

Suitable Grade Level

Grades 11-12