Foundations of Discrete Mathematics | Assessment

MAS162: Foundations of Discrete Mathematics
Semester 2, 2019
ASSIGNMENT 3
Question 1
% Part a Matlab program to check true or false statement for the given values
x=10^(15);
y=2*10^(-2);
z=pi*10^(-17);
a=1;
b=10^(20);
disp(x-1/x==x)
disp(-3+y==-3-y)
disp(3-z==3+z)
disp((a-b)*(a+b)==a^2)
disp((a-b)*(a+b)==b^2)
%Output where 1 means true and 0 means false
%Part B
clear
n=33;%Because MATLAb commences at 1 so the vector must be 33 to store 32 elements
for i=1:n
v(i)=i-n; %this implies that v(0)=0, v(1)=1...v(33)=32
x(i)=8^(-(i-1));
y(i)=(-3/x(i)+1/3)+(1/3+3/x(i));
end
plot(v,y)% since vector contains all the elements 0, 1, 2...32
title('part B')
% output
1
0
1
0

0
Published with MATLAB® R2018a
When the value of i=-16, the value of y changes rapidly as shown in the above MATLAB results.
This is in line with the computer representation of Floating points numbers
Question 2
a.

b.
Question 3
a.
6-3=3 children from the 24 kids are to be invited after inviting her three friends.
This is achievable in (24
3 )=2024 ways
b.
Each of the kids including Lara gets one of T-shirts. Assuming that the kids are arranged
in a line such that only one T-shirt is picked by a kid, the first T-shirt will go to the first
kid and the last one will go to the 7th kid hence the number of combinations
7 !=5040

c.
In this case 5 kids have to be selected from the 7 kids to receive the 5 identical hats. The
number of combinations
( 7
5 ) =21 ways
d.
The word FALAFEL has 3 repeating letters F, L and A and since a child has to pick only
one letter, this combination is not possible hence the probability
0
The word oranges has 7 letters with no repeating letter of the alphabet. It can only be
formed if the first skid pick letter O out of the 26 letters out of which the remaining
letters will be 25, if the next kid picks R, the remaining letters will be 24 and so on hence
probability
1
26 × 1
25 ×… 1
( 26−7+1 ) =19 !
26 !
The word SEATBELT has 8 letters and a repeating letters which can never be formed
hence its probability
0
e.
For the Brown hair kids,
( 3
2 )=3
From the blond hair kids
( 4
3 )=4
Number of combination 3 × 4=12 ways
f. There are seven kids, the number of possible words that can be constructed
7 ×7=49 words
Question 4.
a. This is the probability that a student failed in ENG101 or passed in ENG102. It is the
probability that a student passed ENG102 after failing ENG101.
b. A and B are independent events because the occurrence of A does not affect the
occurrence of B.
c. This is the probability that failed in ENG102 and after passing ENG101.
=0.71*(1-0.6289) =0.263481
d. A and B are not disjoint event. Two events are disjoint if the probability of the two
events occurring at the same time is zero
P(A and B)=0
P(A B)= 0.71*0.6289=0.446519

Comment: the solution has been provided based on the instructions in the original pdf
document.