2. P(X = x) =55 x², if x € {1,2,3,4and 5} %3D 0,otherwise 3. P(X = x) = if x € {1,2,3,4,5,6} 0,otherwise if x € {1,2,3, 4,5, 6) 0, otherwise 4. P(X = x) =91
2. P(X = x) =55 x², if x € {1,2,3,4and 5} %3D 0,otherwise 3. P(X = x) = if x € {1,2,3,4,5,6} 0,otherwise if x € {1,2,3, 4,5, 6) 0, otherwise 4. P(X = x) =91
A First Course in Probability (10th Edition)
10th Edition
ISBN:9780134753119
Author:Sheldon Ross
Publisher:Sheldon Ross
Chapter1: Combinatorial Analysis
Section: Chapter Questions
Problem 1.1P: a. How many different 7-place license plates are possible if the first 2 places are for letters and...
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Note: Number 1 is already answered which is on the right picture. On the right picture should be the format for your answer in number 2, 3, and 4.
Transcribed Image Text:WEEK 2
Show that each of the
Allowing is o probobility mass Runchon
4) P(X-x) =3X, f xE (L2)
0, otherwise
By Subshhuhion
P(X= 4)=
P(X = 2)
Therfove P(x) 30
Check the sum of Probalaliies
+专
2.
%3D
3.
PMF
Transcribed Image Text:Weekly activity (WEEK2)
Show that each of the following is a probability mass function.
if x E (1,2}
1. P(X = x) =
0, otherwise
=x2, if x E {1,2,3,4and 5}
0, otherwise
2. P(X = x) ={55
if x € {1,2,3, 4,5,6)
0, otherwise
3. P(X = x) = {6'
if x € {1,2,3,4,5, 6}
4. P(X = x) = 91'
0, otherwise
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