Probability Theory Assignment: Problems, Solutions, and Analysis

This assignment presents a series of probability theory problems with detailed solutions. The problems cover various concepts, including probability mass functions, cumulative distribution functions, discrete and continuous distributions, and the application of integration and summation techniques. The solutions provide step-by-step explanations, addressing issues such as incorrect integration limits and the application of different distribution types. The assignment also includes analysis of the properties of different functions and distributions, such as determining whether they satisfy the conditions for being a distribution function or density function, and calculating probabilities based on given conditions and areas under curves. The document emphasizes the importance of understanding the underlying principles and applying the correct mathematical tools to solve probability-related problems. It offers a comprehensive guide for students to learn and understand probability theory, with a focus on clear explanations and practical application.

Probability Theory
Problem 1
The probability mass function p(x)= p1 +P2 +P3 +P4 +………+Pn=1
=c + 1
8 c + 1
27 c + 1
64 c +……..+ 1
n3 =1
=c+0.125c +0.03704c + 0.015625c +…………… 1
n3 =1
C(1 +0.125 +0.03704 +0.015625c+……… 1
n3 )=1
C =
1
1+ 0.125+0.03704+ 0.015625 c+… … … 1
n3
Problem 2
fx(x)=P[X=x]=
{ 1 for x=1
1
8 for x=2
1
27 for x=3
1
64 for x=4
… ..
… ..
0 otherwise
The cumulative distribution function:

Fx(x)=P[X ≤x]=∑
n =1
[ x]
P [ X =n ] =
{ 0 for x <1
1
8 c for 1 ≤ x ≤ 2
35
216 c for 2≤ x ≤ 3
2275
13824 c for 3 ≤ x ≤ 4
… …..
… …..
1 for n ≤ x
Fx(2)=
{ 0 for x <1
1
8 c f 0 r 1≤ x<2
1 for 2<x ≤ 2
Problem 3
P1=1
1 =1 ,P2= 1
1=1, P=P1xP2=1
Given P,p1,p2 such that P=P1 x P2
P[{(x,y):y>2x }]=P[{(0x1),(1x2):(1x2)>(2x0x1)}]
=P[{(0,2):2>0}}
=1
Px[x]=P[X=x]=
{ 1 for x ∈[0,2]
p 1
p for 0 ≤ x ≤1
p 2
p for 1 ≤ x ≤2
… .
0 otherwise

Problem 4
P1=1
1=1 ,P2
1
1 =1 P=P1 xP2=1
P[{(x,y):y>2x}]=P[{(0*1, 1
2*3
2,1*2):2>2*0*1)}]
=P[{0,2,2):2>0)}]
=1
Px[{(x,y):y>2x)}]=
{1 for 0 ≤ x ≤ 2
p 2
p
p 1
p
0 otherwise
Problem 5
P1= 1
b−a = 1
1=1 ={
1 for a ≤ x ≤ b
…..
0 otherwise
,P2¿∫
0
∞
xdx 1
b−a+1 = 1
3 =
{ 1
3 for a ≤ x ≤ b
… …
0 for x <a , x <b
P=P1
xP2 =1
3*1 =1
3
P({x,y):y>2x}]= 1
3*1 =1
3
Px({x,y):y>2x}]=1
2 { 1
3 for 0≤ a ≤ b
… .
0 otherwise

Problem 6
Area of the whole triangle = 1
2* 3 *2=3 squared units.
Area of the triangle upto x=1= 1
2*1*1 =1
2
{F1(1)|P1B}= F 1(1)
P 1(B) = e−0.5−e−3
1(3−0.5) =0.6065−0.04979
2.5 =0.5567
F1(1)={0.5567 for t ≠ 0
… …
1 for t =0
Problem 7
P1= 1
b−a =1
2 P2=∫
0
2
∫
0
∞
∫
x− y
∞
1
2 e-(y+z)dzdydx +∫
0
2
∫
0
∞
∫
x− y
∞
1
2 e-(y+z)dzdydx 1
2 e−2
{ 1
2 e0
}{ 1
2 e0 }
=0.06767 ,P3=
∫
0
2
∫
0
∞
∫
x− y
∞
1
2 e−( y + z ) dzdydx+∫
0
2
∫
0
∞
∫
x− y
∞
1
2 e−( y+ z)dzdydx 1
2 e−2
{ 1
2 e0
}{ 1
2 e0 }=0.06767
P=P1 * P2 * P3 =0.5 *0.06767 *0.06767 =0.002290
=0.00229
P{(x,y,z):y+z>x})= {
0.00229 for a≤ x ≤ b
… ..
0 otherwise
Problem 8
Yes

Problem 9
( 2,5)
Problem 10
1, 2 & 4