Probability and Statistics Homework: Numerical Solutions Provided
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Homework Assignment
AI Summary
This document contains the solutions to a probability and statistics homework assignment. The assignment includes 15 questions covering various topics in probability and statistics, such as calculating probabilities of coin flips, combinations, conditional probabilities, expected values, and variance. Each question is answered with a numerical value, providing a comprehensive set of solutions for students to understand and learn from. The solutions include detailed steps and calculations, making it easier for students to follow the logic and grasp the concepts. Additionally, the assignment adheres to the requirement of providing three decimal places in the numerical answers where applicable, ensuring precision and accuracy in the solutions. A bibliography is also included, citing relevant sources used in the assignment.

Running head: PROBABILITY AND STATISTICS
Probability and Statistics
Name of the student:
Name of the University:
Author note:
Probability and Statistics
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Name of the University:
Author note:
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1
PROBABILITY AND STATISTICS
Table of Contents
Answer to the question 1............................................................................................................2
Answer to the question 2............................................................................................................2
Answer to the question 3............................................................................................................2
Answer to the question 4............................................................................................................3
Answer to the question 5............................................................................................................3
Answer to the question 6............................................................................................................3
Answer to the question 7............................................................................................................4
Answer to the question 8............................................................................................................4
Answer to the question 9............................................................................................................4
Answer to the question 10..........................................................................................................5
Answer to the question 11..........................................................................................................5
Answer to the question 12..........................................................................................................5
Answer to the question 13..........................................................................................................6
Answer to the question 14..........................................................................................................6
Answer to the question 15..........................................................................................................7
Bibliography...............................................................................................................................8
PROBABILITY AND STATISTICS
Table of Contents
Answer to the question 1............................................................................................................2
Answer to the question 2............................................................................................................2
Answer to the question 3............................................................................................................2
Answer to the question 4............................................................................................................3
Answer to the question 5............................................................................................................3
Answer to the question 6............................................................................................................3
Answer to the question 7............................................................................................................4
Answer to the question 8............................................................................................................4
Answer to the question 9............................................................................................................4
Answer to the question 10..........................................................................................................5
Answer to the question 11..........................................................................................................5
Answer to the question 12..........................................................................................................5
Answer to the question 13..........................................................................................................6
Answer to the question 14..........................................................................................................6
Answer to the question 15..........................................................................................................7
Bibliography...............................................................................................................................8

2
PROBABILITY AND STATISTICS
Answer to the question 1
The number of flipping a coin is 5
Total number of outcome= 32
(H,T) (H,T) (H,T) (H,T) (H,T)
Hence the required probability two heads out of five flips= 4/32
= 0.125
Answer to the question 2
Number of letters = 6
3 letters string cab be constructed by using 3 letters = 63
= 216
Answer to the question 3
Given that
X= 1,2,3,4,5
P(X)= 1/5
P (X ≤ 3 ¿= P (X=1) +P (X=2) + P (X=3)
= 1/5+1/5+1/5
= 3/5
= 0.6
PROBABILITY AND STATISTICS
Answer to the question 1
The number of flipping a coin is 5
Total number of outcome= 32
(H,T) (H,T) (H,T) (H,T) (H,T)
Hence the required probability two heads out of five flips= 4/32
= 0.125
Answer to the question 2
Number of letters = 6
3 letters string cab be constructed by using 3 letters = 63
= 216
Answer to the question 3
Given that
X= 1,2,3,4,5
P(X)= 1/5
P (X ≤ 3 ¿= P (X=1) +P (X=2) + P (X=3)
= 1/5+1/5+1/5
= 3/5
= 0.6
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PROBABILITY AND STATISTICS
Answer to the question 4
Given that
f (X) = c 0≤ X ≤ 8
= 0 elsewhere
P (X < 7) =?
∫ f ( x ) dx= 1
Or ∫
0
8
c dx = 1
Or c = 1/8
P (X < 7) = ∫
0
7
c dx
= 1/8 (7-0)
= 0.875
Answer to the question 5
fX (x) = x2 0≤ X ≤ 1
P (0.87 ≤ X ≤ 0.94) = ∫
0.87
0.94
x 2 dx
= 1/3 (0.831- 0.658)
= 0.057
Answer to the question 6
Out of 4 flip the total number of outcome = 16
PROBABILITY AND STATISTICS
Answer to the question 4
Given that
f (X) = c 0≤ X ≤ 8
= 0 elsewhere
P (X < 7) =?
∫ f ( x ) dx= 1
Or ∫
0
8
c dx = 1
Or c = 1/8
P (X < 7) = ∫
0
7
c dx
= 1/8 (7-0)
= 0.875
Answer to the question 5
fX (x) = x2 0≤ X ≤ 1
P (0.87 ≤ X ≤ 0.94) = ∫
0.87
0.94
x 2 dx
= 1/3 (0.831- 0.658)
= 0.057
Answer to the question 6
Out of 4 flip the total number of outcome = 16
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4
PROBABILITY AND STATISTICS
= (1*0.12)+(2*0.12)+ (3*0.12)+(4*0.12)/16
= 0.081
Answer to the question 7
Mean = 2
Standard deviation = 2
P (X ≤8) = P ( Z ≤ 3)
Z =( 8- 2) /2 = 3
Answer to the question 8
P(Xi) = 1/6 i = 1,2,3,4,5,6
Even cases = 2,4,6
Pi = 1/3 i=2,4,6
E (X/ A=2,4,6) = (X/A=2) P (A=2)+ (X/A=4) P(A=4) + (X/A=6) P(A=6)
= 2 * 1/3 + 4 *1/3 + 6* 1/3
= 12/3
=4
Answer to the question 9.
Total number of outcome= 6*6*6*6*6
= 7776
No. of different outcomes= factorial (6) = 720
P = 724/ 7776
PROBABILITY AND STATISTICS
= (1*0.12)+(2*0.12)+ (3*0.12)+(4*0.12)/16
= 0.081
Answer to the question 7
Mean = 2
Standard deviation = 2
P (X ≤8) = P ( Z ≤ 3)
Z =( 8- 2) /2 = 3
Answer to the question 8
P(Xi) = 1/6 i = 1,2,3,4,5,6
Even cases = 2,4,6
Pi = 1/3 i=2,4,6
E (X/ A=2,4,6) = (X/A=2) P (A=2)+ (X/A=4) P(A=4) + (X/A=6) P(A=6)
= 2 * 1/3 + 4 *1/3 + 6* 1/3
= 12/3
=4
Answer to the question 9.
Total number of outcome= 6*6*6*6*6
= 7776
No. of different outcomes= factorial (6) = 720
P = 724/ 7776

5
PROBABILITY AND STATISTICS
= 0.093
Answer to the question 10
f(x) = 1/6 0≤ X ≤ 6
E(X)= ∫
0
6
xf (x)dx
= 1/ 6∫
0
6
xdx
= 1/6 (1/2) (36)
= 3
Answer to the question 11
Let A, B and C denote the event that first, second and third component works.
Given that
P(A)= 0.2
P (B) =0.5
P(C) = 0.3
Hence the probability that the whole system works = P(A)P(B)P(C)
= 0.2*0.5*0.3
= 0.03
Answer to the question 12
Given that
fX(x)= C x2 -4≤ X ≤ 4
PROBABILITY AND STATISTICS
= 0.093
Answer to the question 10
f(x) = 1/6 0≤ X ≤ 6
E(X)= ∫
0
6
xf (x)dx
= 1/ 6∫
0
6
xdx
= 1/6 (1/2) (36)
= 3
Answer to the question 11
Let A, B and C denote the event that first, second and third component works.
Given that
P(A)= 0.2
P (B) =0.5
P(C) = 0.3
Hence the probability that the whole system works = P(A)P(B)P(C)
= 0.2*0.5*0.3
= 0.03
Answer to the question 12
Given that
fX(x)= C x2 -4≤ X ≤ 4
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PROBABILITY AND STATISTICS
∫
x
❑
f ( X ) dx=¿ ¿ 1
Or ∫
−4
4
C x 2=¿ ¿ 1
Or C (X3/3)-44= 1
Or (128 C)/3 = 1
Or C = 1/42.66
= 0.023
Answer to the question 13
Hence the required solution is c3
10
Numerator = fac(10)
Denominator = fac(3) * fac(7)
Thus c3
10 = 10*9*8*fac(7)/ 3*2*1*fac(7)
= 120
Answer to the question 14
Number of flip =4
Head in a single flip =0.3
Total number of outcome= 16
Let X denote the flip
X=1, all the flip are same
0, otherwise
PROBABILITY AND STATISTICS
∫
x
❑
f ( X ) dx=¿ ¿ 1
Or ∫
−4
4
C x 2=¿ ¿ 1
Or C (X3/3)-44= 1
Or (128 C)/3 = 1
Or C = 1/42.66
= 0.023
Answer to the question 13
Hence the required solution is c3
10
Numerator = fac(10)
Denominator = fac(3) * fac(7)
Thus c3
10 = 10*9*8*fac(7)/ 3*2*1*fac(7)
= 120
Answer to the question 14
Number of flip =4
Head in a single flip =0.3
Total number of outcome= 16
Let X denote the flip
X=1, all the flip are same
0, otherwise
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7
PROBABILITY AND STATISTICS
P(X=1)= 2/16
= 0.125
Answer to the question 15
Given
X= 4,5,6,7,8,9,10
N= 7
P = 1/7
E(X)= 1/7( 4+5+6+7+8+9+10)
= 7
E(X^2) = 1/7( 4^2+5^2+6^2+7^2+8^2+9^2+10^2)
= 371/7
= 53
Var (X) = 53- 49
= 4
PROBABILITY AND STATISTICS
P(X=1)= 2/16
= 0.125
Answer to the question 15
Given
X= 4,5,6,7,8,9,10
N= 7
P = 1/7
E(X)= 1/7( 4+5+6+7+8+9+10)
= 7
E(X^2) = 1/7( 4^2+5^2+6^2+7^2+8^2+9^2+10^2)
= 371/7
= 53
Var (X) = 53- 49
= 4

8
PROBABILITY AND STATISTICS
Bibliography
Li, N., Liu, X., Xie, W., Wu, J., & Zhang, P. (2013). The return period analysis of natural
disasters with statistical modeling of bivariate joint probability distribution. Risk
Analysis: An International Journal, 33(1), 134-145.
Man'ko, M. A. (2013). Joint probability distributions and conditional probabilities in the
tomographic representation of quantum states. Physica Scripta, 2013(T153), 014045.
PROBABILITY AND STATISTICS
Bibliography
Li, N., Liu, X., Xie, W., Wu, J., & Zhang, P. (2013). The return period analysis of natural
disasters with statistical modeling of bivariate joint probability distribution. Risk
Analysis: An International Journal, 33(1), 134-145.
Man'ko, M. A. (2013). Joint probability distributions and conditional probabilities in the
tomographic representation of quantum states. Physica Scripta, 2013(T153), 014045.
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