AM0S05 Portfolio Assignment: Probability and Statistics Solutions

This document presents a student's solutions to a probability and statistics portfolio assignment, specifically for the AM0S05 module. The assignment covers several key concepts in probability, including calculating probabilities of events, determining whether events are mutually exclusive or exhaustive, and applying probability formulas. The solutions address problems involving random variables, conditional probability, and the use of replacement in probability calculations. The student demonstrates an understanding of basic probability principles, providing step-by-step solutions to the problems. References to probability textbooks are also included. The assignment is a part of the AM0S05 module and contributes 20% to the overall grade, with a submission deadline of February 28, 2020. The document provides a comprehensive approach to solving probability problems with calculations and explanations.

Running head: Maths
Maths
Name of the Student
Name of the University

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Table of Contents
Q1...............................................................................................................................................3
Q2...............................................................................................................................................3
Q3...............................................................................................................................................4
Q4...............................................................................................................................................5
Q5...............................................................................................................................................5
References:.................................................................................................................................6

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Q1.
Here α=10
P(A) * P(B) =
7
10∗13
15 = 91
150 ≠ 31
45 =P ( A ∩ B )
Thus, A and B are not exhaustive.
Q2.
There are 10 balls numbered 1 to 10.
E1 = "a ball with a odd number is extracted"
E2 = "a ball with a multiple of 3 is extracted"
E3 = "a ball with a even number is extracted"
E4 = "a ball with a number less than 10 is extracted"
a)
P (E1) = 5
10 =1
2 = 0.5
P (E2) = 3
10 = 0.3
P (E3) = 5
10 = 0.5
P (E4) = 9
10 = 0.9
b)
P ( E 1∩ E 2 )= 2
10 = 1
5 = 0.2

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P ( E 1∪ E 2 )=P ( E 1 ) + P ( E 2 ) −P ( E 1 ∩ E 2 )= 1
2 + 3
10 − 1
5 =0.6
c)
P ( E 1∩ E 2 )=0.2 and P ( E 1 )∗P ( E 2 )=0.5∗0.3=0.15
Thus P ( E 1∩ E 2 ) ≠ P ( E 1 )∗P ( E 2 )∧are not mutually exclusive .
d)
P ( E 3∩ E 4 ) = 4
10 =0.4
P ( E 3∪ E 4 ) = P (E3) + P (E4) - P ( E 3∩ E 4 )=¿0.5+0.9 - 0.4 = 1
e)
No E3 and E4 are not mutually exclusive.
f)
Yes E1 and E3 are mutually exclusive.
g)
No, (E1 ∩ E2) and (E3 ∩ E4) are not mutually exclusive as 6 is common.
Q3.
P(A) = 2/10 = 0.2, P(B) = 3x, P(A ∪B) = 6x, P(A ∩ B) = x
P(A ∪B) = P(A) + P(B) – P(A∩ B ¿
6x = 0.2 + 3x – x
X= 0.2/4 = 0.05

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Q4.
A urn contains 10 white balls and 10-2 = 8 black balls. Two balls are selected at random
with replacement
Probability that
a) both balls are white = 10C 2
18C 2 =0.29
b) both balls are black = 8C 2
18C 2 =0.18
c) the balls are of the same colour: 0.18 +0.29 = 0.47
d) the balls are of different colours: 1-0.47 = 0.53
e) both balls are black given that they are of the same colour: 0.18/0.47 = 0.38
f) both balls are white given that they are of the same colour: 0.29/0.47 = 0.62
Q5.
Without replacement:
a) both balls are white =
10
18∗9
17 =0.29
b) both balls are black =
8
18∗7
17 =0.18
c) the balls are of the same colour: 0.18 +0.29 = 0.47
d) the balls are of different colours: 1- 0.47 = 0.53
e) both balls are black given that they are of the same colour: 0.18/0.47 = 0.38
f) both balls are white given that they are of the same colour: 0.29/0.47 = 0.62

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References:
Ross, S.M., 2014. A first course in probability (Vol. 7). Upper Saddle River, NJ: Pearson
Prentice Hall.
Keynes, J.M., 2016. A treatise on probability. Courier Corporation.

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