Probability Problems and Solutions: Solved with Step-by-Step Guide

This assignment provides detailed solutions to a range of probability problems, starting with basic probability calculations using the rule of addition, including scenarios with complementary events. It then delves into conditional probability, demonstrated through problems involving students with siblings. The assignment also explores probability with dice, calculating probabilities of specific outcomes and identifying values based on frequency. Further, it applies Bayes' Theorem to calculate conditional probabilities and examines the concepts of intersection and union of events. The assignment also uses probability tree diagrams to solve complex probability questions. The problems include finding the probability of being late using different routes and analyzing meeting probabilities among individuals going to different locations. Desklib offers more solved assignments and study resources for students.

Question 1
Using the rule of addition
P ( A ∪ B ¿=P ( A ) + P ( B ) −P( A ∩ B)
Where
P ( A ) =1−P ( A )
P ( A )=1− 1
4
P ( A )= 3
4
P ( B )= 1
2
P ( A ∩B )= 1
3
Substituting in the original equation
P ( A ∪ B ¿=3 /4+ 1/2−1/3
P ( A ∪ B ¿=11 /12
Question 2
Using the conditional probability
P(Total Children=3|students with 2 siblings)
Total number of students=30
Number of students with two siblings=9
P(Total Children=3|students with 2 siblings)=9/30 or 3/10
Question 3
Assuming each dice has 6 sides
We can therefore answer the resulting questions

(a). Possibility Space Diagram
Dice 1
1 2 3 4 5 6
1 1 2 3 4 5 6
2 2 4 6 8 10 12
3 3 6 9 12 15 18
Dice 2 4 4 8 12 16 20 24
5 5 10 15 20 25 30
6 6 12 18 24 30 36
(b). P(10)=2/36=1/18
(c). P(25)=1/36
(d). P(27)=0/36=0
(e). P(t)=1/9=4/36
Which value “t” has a frequency of 4 in the possibility space diagram?
6 appears four times in the possibility space diagram and it has a probability of 1/9
Then t=6
Question 4
(a). Using Bayes’ Theorem
P ( A |B ) = ( P ( B| A )∗P ( A ) }
P ( B )
Hence
P ( B| A ) = ( P ( A |B )∗P ( B ) }
P ( A )
P ( B| A ) =
2
3∗1
3
2
3

P ( B|A )= 1
2
(b). Using the notation
P ( B| A ) =P( B ∩ A)/ P ( A )
Hence
P ( B ∩ A )=P ( B|A )∗P ( A )
P ( B ∩ A )=
1
2∗2
3
P ( B ∩ A )= 2
6
Using the rule of addition
P ( B ∪ A ) =P ( B ) + P ( A ) −P ( B ∩ A )
P ( B ∪ A ) =1
3 + 2
3 − 2
6
P ( B ∪ A ) = 4
6
(b) Summing the intersection and union of A and B
A and B are exhaustive because P ( B ∩ A ) + P ( B ∪ A ) =1
Question 5
Home
D E
Early Late Early Late
0.70.3
0.20.80.70.3

(a). The probability that am late
P(L)=P(Late with E)*P(Late with D)
P(L)=0.2*0.7
P(L)=0.14
(b). The probability that am late and I take route E
P(E|L)=0.7
Question 6
(a). Possibility Tree diagram
Brian
Charlie
Ali
Origins Mungo’s K-bar Not
0.5 0.5
0.5 0.5
0.5 0.5

Find probability
(b). P(Ali and Brian meet)=0.5*0.5=0.25
(c). P(Charlie and Brian meet)=0.5*0.5=0.25
(d). P(all meet)=0
(e). P(all go to different places)=(0.5*0.5*0.5)+(0.5*0.5*0.5)=0.25
(f). P(at least 2 meet)=(0.5*0.5)+(0.5*0.5)=0.5