Let X₁, X2 be 2 mutually independent discrete random variables. The first variable X₁ can be a whole number from 1 to 5. Its distribution function is 6 - i 15 P(X₁ = 4) = P(X₂ = 6) = 1/6 P(X₁ = 4 and X₂ 2 The second variable X₂ can be a whole number from 1 to 6. It has a uniform distribution. First, find the exact values (using fractions) of = 2/15 :6) = --- --- = (2/15)(1/6 m₁(i) = Find the probability that Y = 2, P(Y = 2) = Hint: Use cases based on what X₁ and X₂ are. Let Y denote the maximum of the X;'s. Find the probability that Y = 1: P(Y = 1) =

Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter10: Sequences, Series, And Probability
Section10.8: Probability
Problem 32E
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Just need help solving for P(Y=1) and P(Y=2).

Let X₁, X₂ be 2 mutually independent discrete random variables.
The first variable X₁ can be a whole number from 1 to 5. Its distribution function is
6 - i
15
P(X₁ = 4) =
P(X₂ = 6) = 1/6
2
The second variable X₂ can be a whole number from 1 to 6. It has a uniform distribution.
First, find the exact values (using fractions) of
= 2/15
P(X₁ :
=
= 4 and X₂ = 6) = (2/15)(1/6
m₁ (i) =
●‒‒
=
Let Y denote the maximum of the X; 's. Find the probability that Y = 1:
P(Y = 1) =
Find the probability that Y
P(Y = 2) =
Hint: Use cases based on what X₁ and X₂ are.
Transcribed Image Text:Let X₁, X₂ be 2 mutually independent discrete random variables. The first variable X₁ can be a whole number from 1 to 5. Its distribution function is 6 - i 15 P(X₁ = 4) = P(X₂ = 6) = 1/6 2 The second variable X₂ can be a whole number from 1 to 6. It has a uniform distribution. First, find the exact values (using fractions) of = 2/15 P(X₁ : = = 4 and X₂ = 6) = (2/15)(1/6 m₁ (i) = ●‒‒ = Let Y denote the maximum of the X; 's. Find the probability that Y = 1: P(Y = 1) = Find the probability that Y P(Y = 2) = Hint: Use cases based on what X₁ and X₂ are.
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