Probability and Statistics Assignment - Semester 1, University

This assignment covers key concepts in probability and statistics. Part 1 focuses on probability, including urn problems, calculating probabilities of different events, and analyzing the number of times an event occurs. Part 2 addresses mathematical statistics, including calculating the mean and variance, exploring probability distributions, and graphing functions. The assignment explores various scenarios, providing detailed solutions and calculations for each problem. The solutions include graphical representations and the application of relevant formulas to arrive at the correct answers. The assignment addresses a range of statistical techniques, including the analysis of probability distributions and the application of statistical methods to solve complex problems. The solutions are presented in a clear and concise manner, making it easy to understand the steps involved in solving each problem.

Theory of Probability and Mathematical Statistics
Name of Student
Name of Professor
Date

PART A
Theory of Probability
Question 1
First urn
5 white, 4 black balls, 1 ball taken
Second urn
7 white, 4 black balls, 4 balls taken
a) Probability that all the balls taken are of the same color.
¿ P(all white)∨P(all black )
P(all white)= 5
9 × 7
11 × 7
11 × 7
11 × 7
11
¿ 0.0911
P ( all black )= 4
9 × 4
11 × 4
11 × 4
11 × 4
11
¿ 0.0777
Thus,
P ¿
¿ 0.1688
b) Probability that only three balls are white
¿ P ( WWWB B )∨P (BWWWB )

P ( WWWBB )= 5
9 × 7
11 × 7
11 × 4
11 × 4
11
¿ 0.0297
P( BWWWB= 4
9 × 7
11 × 7
11 × 7
11 × 4
11 )
¿ 0.0416
Thus, probability that only three balls are white,
¿ 0.0297+ 0.0416
¿ 0.0713
c) Probability that at least one ball is white.
¿ 1−P(all black)
¿ 1−0.0777
¿ 0.9223
Question 2
The reliability of the device is given by;
¿ p4 ×(1− p1 )2 × p3 ׿
¿ 0.7 ×(1−0.8)2 ×0.6 × ¿
¿ 0.7 × 0.04 ×0.6 × 0.058
¿ 0.0009744
Question 3
There are 6 possible scenarios for the initial balls in the urn

¿ P ( WWWWW ) ∨P ( WWWWB ) ∨P ( WWWBB ) ∨P ( WWBBB ) ∨P ( WBBBB ) ∨P (BBBBB)
Adding 4 white balls to these scenarios we have the probability of removing 4 white
balls as
¿ ( 1
9 × 1
9 × 1
9 × 1
9 )+ ( 1
8 × 1
8 × 1
8 × 1
8 )+( 1
7 × 1
7 × 1
7 × 1
7 )+( 1
6 × 1
6 × 1
6 × 1
6 )+ ( 1
5 × 1
5 × 1
5 × 1
5 )+( 1
4 × 1
4 × 1
4 × 1
4
¿ 0.0709
Question 4
There are 3 possibilities for the balls taken from the first urn.
¿ P ( WWW )∨P (WWB ) ∨P (WBB)
The probability that all the balls taken from the second urn are all white for all the
possibilities is thus given by;
¿ ( 7
11 × 7
11 × 7
11 )+ ( 6
11 × 6
11 × 6
11 )+( 5
11 × 5
11 × 5
11 )
¿ 0.2577+0.1623+ 0.0939
¿ 0.5139
Question 5
a) Exactly one time
¿ ( n
x ) px .(1− pn− x)
P( X=1)=( 480
1 ) 0.00421 .(1−0.0042480−1 )
¿ 0.2 685

b) The most probable number of times
P( X=480)= ( 480
480 ) 0.0042480 .(1−0.0042480−480)
¿< 0.0000001
≈ 0
c) Less than 12 times
P ( X <12 )=0.99999865583≈ 1
d) More than two times
P ( X >2 ) =0.3277
Question 6
a) Happen exactly 223 times.
P ( X=223 ) =( 540
223 ) 0.44223 .(1−0.44540−223)
¿ 0.01558
b) More than 184 times but less and less than 244 times
P ( 184< X < 244 ) =P ( X >184) × P( X <224)
P ¿
P( X<224)=0.11060281506
Thus,
P ( 184<X <244 ) =0.99999837588× 0.11060281506
¿ 0.1106
PART TWO

Mathematical Statistics
Question 1
X N−1 N N +1 N +2
P 0.015 0.25 0.355 0.38
X2 (N −1)2 N2 (N +1)2 ¿ ¿
The mean
M ( x ) =∑ XP( x )
¿ 0.015 ( N−1 ) +0.25 N +0.355 ( N +1 ) +0.38 (N +2)
¿ 0.015 N +0.25 N +0.355 N +0.38 N−0.015+0.355+0.76
¿ N +1.1
Variance
D ( x ) =∑ X2 P( x)−∑ XP(x )
∑ X2 P( x )=0.015 ( N2−2 N +1 ) +0.25 N +0.355 ( N2 +2 N +1 ) +0.38( N 2+4 N + 4)
¿ 0.015 N2−0.03 N +0.015+ 0.25 N + 0.355 N2 +0.71 N +0.355+0.38 N2 +1.52 N + 1.52
¿ 0.015 N2 +0.355 N2+ 0.38 N 2−0.03 N +0.25 N +0.71 N +1.52 N +0.015+0.355+ 1.52
¿ 0.75 N2 +2.45 N + 1.89
∑ XP(x)=N + 1.1
Thus,
D ( x ) =0.75 N2+ 2.45 N +1.89− ( N +1.1 )
¿ 0.75 N2 +1.45 N −0.79

σ ( x ) = √ D ( x )
¿ √ 0.75 N2+1.45 N −0.79
Polygon of distribution
1
0
0.05
0.1
0.15
0.2
0.25
0.3
0.35
0.4
Chart Title
N-1 N N+1 N+2
Function
F ( x )=
{0.015 , x=N−1 ,
0.25 , x =N ,
0.355 , x=N +1
0.38 , x=N +2

Graph
N-1 N N+1 N+2
0
0.05
0.1
0.15
0.2
0.25
0.3
0.35
0.4
Chart Title
Question 2
i) Law of probability distribution
X 0 1 2 3 4
P 0 0.5 0.125 0.0625 0.03125
ii) F(x) and graph
F ( x ) = { 0 , x=0 ,
0.5x , 1 ≤ x ≤ 4

0.5 1 1.5 2 2.5 3 3.5 4 4.5
0
0.1
0.2
0.3
0.4
0.5
0.6
Chart Title
Question 3
F ( x )=
{ 0 , x ≤ 1 ,
1
2 ( x−1 ) ,1< x ≤3
1, x >3
i) Numerical characteristics
M ( x ) =∫
−∞
∞
xF ( x ) dx
∫
1
3
x . 1
2 ( x−1 ) dx= 1
2 ( x3
3 − x2
2 )∨¿1
3 ¿

¿ 1
2 (5−−0.167)a
¿ 2.5835
D ( x ) =∫
−∞
∞
¿ ¿ ¿
∫
−∞
∞
x2 F ( x ) dx=¿∫
1
3
x2 . 1
2 ( x−1 ) dx ¿
¿ 1
2 ( x4
4 − x3
3 )∨¿1
3 ¿
¿ 1
2 (11.25−−0.0833)
¿ 11.1667
Thus,
D ( x ) =11.1667−6.6744 ¿ 4.4923
σ ( x ) = √ D ( x )
¿ √4.4923
¿ 2.1195

ii) f ( x ) =
{ 0 , x ≤1 ,
x2
4 − x
2 , 1< x ≤ 3 ,
x x >3
Graph of F(x)
-2 -1 0 1 2 3 4 5 6
0
0.2
0.4
0.6
0.8
1
1.2
Chart Title
Graph of f(x)
x -1 0 1 2 3 4 5
f(x) 0 0 0 0 0.75 4 5

-2 -1 0 1 2 3 4 5 6
0
1
2
3
4
5
6
Chart Title
iii) Between -1 and 2
f ( x )=0+ 1
4
¿ 1
4
Thus, P(-1,2) = 1
4