Given a p.m.f. of x as below: (e-33x f(x) x = 0, 1, 2, ... = x! 0 O. W and y = 2x, then the p.m.f. of x of y is: e g(y)= y = 0, 2, 4, ... 0. W y = 0, 2, 4,... 0. W Option 3 g(y) = y! 0 (e-³3² 0

A First Course in Probability (10th Edition)
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ISBN:9780134753119
Author:Sheldon Ross
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Chapter1: Combinatorial Analysis
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Given a p.m.f. of x as below:
f(x)=
(e-33x
x!
x = 0, 1, 2, ...
0
0. W
and y= 2x, then the p.m.f. of x of y is:
g(y)=
y = 0, 2, 4, ...
0. W
y = 0, 2, 4,...
0. W
Option 3
g(y)
&
0
Option 1
0
g(y) =
O Option 2
e-³32
0
y = 0, 1, 2,...
0.W
Transcribed Image Text:Given a p.m.f. of x as below: f(x)= (e-33x x! x = 0, 1, 2, ... 0 0. W and y= 2x, then the p.m.f. of x of y is: g(y)= y = 0, 2, 4, ... 0. W y = 0, 2, 4,... 0. W Option 3 g(y) & 0 Option 1 0 g(y) = O Option 2 e-³32 0 y = 0, 1, 2,... 0.W
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