Dec 1998 MA214: Discrete Mathematics (Question 1)

(a)

Suppose that A and B are the sets defined below.

A = {x | x is a real number such that x² = 1}
B = {x | x is an integer such that x² = 1}

(i) List the members of set A and write out in full the power set P(A).

[2]

1 mark for A = {1, -1}
1 mark for P(A) = {Ø, {-1}, {1}, {-1, 1}}
Candidates may write {} instead of Ø but not {Ø}.

(ii) What is the cardinality of the power set P(B)? Show all of your working.

[2]

1 mark for B = Ø or {} or cardinality of B is zero.
Candidates do not get the mark if they write B = {Ø}.
1 mark for cardinality of P(B) = 2⁰ = 1.

(b)

Let A be the vector

and B be the vector

. Calculate

(i) AB^t

[2]

1 mark for getting a 3 x 3 matrix. 1 mark for getting the correct matrix.

(ii) A^tB

[2]

1 mark for getting a 1 x 1 matrix. 1 mark for getting the correct matrix.
(-30)

(c)

Calculate the value of

[2]

1 mark for knowing that the limit of the right hand side is infinity.
1 mark for getting that the limit of the whole expression is infinity.

(d)

Let the sets P, O and I be the sets described below.

P    The set of all prime numbers
O   The set of all odd numbers
I    The set of all integers

Taking the real numbers to be the universal set, draw a Venn Diagram showing the relationships between P, O and I.

[3]

1 mark for having P and O as subsets of I.
1 mark for O and I intersecting.
1 mark for having regions for odd numbers which are not primes and for primes which are not odd numbers.

(e)

Two classes, each containing 30 students, sit the same examination. The means and variances are given below.

	Mean	Variance
Class A	67.3	5.9
Class B	73.4	18.2

Comment on the relative abilities of the two classes justifying your observations by referring to both the means and the variances.

[3]

1 mark for knowing that on average Class B performed better.
1 mark for knowing that the spread of abilities was greater in Class B.
1 mark for giving the respective justifcations that this is because the mean is greater.

(f)

The probability that an event A occurs is 0.6
The probability that an event B occurs given that event A has occurred is 0.8.
The probability that event B occurs given that event A has not occurred is 0.3

(i) Calculate the probability of events A and B both occurring.

[1]

P(A

B) = 0.6 x 0.8 = 0.48

(ii) Calculate the probability of event B occurring and event A not occurring.

[1]

B) = 0.4 x 0.3 = 0.12

(iii) Hence calculate the probability of event B occurring.

[1]

P(B) = 0.48 + 0.12 = 0.6

(iv) Hence calculate the probability that event A occurred given that event B occurred.

[1]

P(A | B) = P(A

B)/P(B) = 0.48/0.6 = 0.8