Probability I
Example Sheet Solutions 7
1. This is a binomial question.
(a) Let F denote the number of customers who choose fish. Then F Bin(5, 3/14). Hence,
2
3
5
3
3
P (F = 2) =
1
= 0.223
2
14
14
(b) Let A denote the number of customers who choose fis

MATH10111 - FEEDBACK ABOUT COURSE
TEST
Question 1: This caused very few problems.
Question 2: Remember that each quantifier (in this case x N) and the predicate must each be negated.
Question 3: Caused very few problems.
Question 4: Remember cfw_ is NOT a

MATH10111 - SOLUTIONS TO EXERCISES 10
10.2(i)(a) Yes since x x + 1 for all x R.
(b) No since 2R5 but 5 6 R2.
(c) No since 2R1 and 1R0 but 2 6 R0.
(ii)(a) Yes since x2 0 for all x R.
(b) Yes since xRy xy 0 yx 0 yRx.
(c) No since 1R0 and 0R1 but 1 6 R1.
(ii

MATH10111 - SOLUTIONS TO EXERCISES
WEEK 1
1.2 (i) Proposition. Negation: 210 1 is not a prime number.
(ii) Predicate. Negation: The largest angle of triangle x is greater than /2.
(iii) Predicate. Negation: The real number x is such that x2 x.
(iv) Propos

MATH10111 - SOLUTIONS TO EXERCISES 3
3.3 For n any natural number, write P (n) for the statement 13 divides
2
+ 3n+2 .
P (1) is true since 24+2 + 31+2 = 64 + 27 = 91 = 13 7.
Now suppose that P (r) is true. We will show that P (r + 1) must then also be tru

MATH10111 - SOLUTIONS TO EXERCISES
WEEK 2
2.3 Suppose that the statement is false, and argue for a contradiction. Then there
exist integers a and b such that 24a + 3b2 = 7. Hence 3(8a + b2 ) = 7. Since 8a + b2 is
an integer, we have 3|7, a contradiction.

MATH10111 - SOLUTIONS TO EXERCISES 9
9.1 (i), (ii), (iv) and (v) are true, (iii) and (vi) are false.
9.2 (i) 9; (ii) 9; (iii) 0; (iv) 12; (v) 4; (vi) 1.
9.3 If a is even, then a = 2k for some k Z. Hence a2 = 4k 2 0 mod 4. If a is
odd, then a = 2k + 1 for

MATH10111 - SOLUTIONS TO EXERCISES 5
5.2 (i) True. Let x R. Taking y = 1 x R, we have x + y = x + (1 x) = 1 > 0.
(ii) True. Let x R. Taking y = x 1 R, we have x y = x (x 1) = 1 > 0.
(iii) False. The negation of the statement is x R, y R, x + y 0.
We prove

MATH 10111: SETS, NUMBERS AND
FUNCTIONS - COURSE TEST SOLUTIONS 2011
Question 1 (a):
p
T
T
F
F
q (p q)
T
F
F
T
T
T
F
T
Question 1 (b):
p
T
T
F
F
q p (q)
T
F
F
T
T
T
F
T
Question 2:
x N, x2 < x
Question 3:
D
Question 4:
cfw_, cfw_1, cfw_, cfw_e, cfw_1, ,

MATH10111 - SOLUTIONS TO EXERCISES 7
7.1 |A B| = 4.2 = 8. N8 = cfw_1, 2, 3, 4, 5, 6, 7, 8. Define f : N8 A B by
f (1) = (a, e), f (2) = (b, e), f (3) = (c, e), f (4) = (d, e), f (5) = (a, f ), f (6) = (b, f ),
f (7) = (c, f ), f (8) = (d, f ).
7.2
|A B C|

MATH10111 - SOLUTIONS TO EXERCISES 8
8.1 The number of functions A A is nn and the number of bijections is n!, so
the answer is nn n!
8.2 Write |A| = m and |B| = n. If n < m, then there are no injective
functions
n
A B. Suppose n m. The number of subsets

MATH10111
Two and a Half Hours
THE UNIVERSITY OF MANCHESTER
SETS, NUMBERS AND FUNCTIONS B
? January 2014
?
Answer ALL SIX questions in Section A (40 marks in total). Answer FOUR of the SIX questions
in Section B (60 marks in total). If more than FOUR ques