August 1997
QT211: QUANTITATIVE ANALYSIS FOR MANAGEMENT

QUESTION 2

Total Marks: 20 Marks

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2. (a) A regular octahedron, which has eight triangular faces numbered from 1 to 8, is to be used as a die. The score for a throw is the number on the face which is uppermost when the die is thrown on a horizontal table. If two such dice are thrown, find the probabilities of the following.
There are two ways of approaching this question.
(i) The total score on the two dice is at least 14. [4]
One is to note that the number of possible pairs is 8 x 8 = 64, [1]
and then to count the number of ways obtaining a sum of 14, 15 or 16. This can be achieved by the pairs (6,8), (8,6), (7,7), (7,8), (8,7), or (8,8) [2]
so that the probability of obtaining a total score of at least 14 is 6/64 = 3/32. [1]
Alternatively, we can work out the probability of any particular pairs occurring as 1/8 X 1/8 = 1/64, and there are six such pairs (6,8), (8,6), (7,7), (7,8), (8,7), or (8,8) [2]
so that the probability of obtaining a total score of at least 14 is 6/64 = 3/32. [1]
[4 marks]
(ii) The difference between the scores on the two dice is two. [4]
To obtain a difference of two we can obtain any of the pairs (1,3), (2,4), (3,5), (4,6), (5,7), (6,8), [2]
or in the reverse order, giving a total of 12 such pairs, [1]
and an overall probability of 12/64 = 3/16. [1]
[4 marks]
(b) Two players A and B throw both dice in turn, A throws first. The first player to obtain a difference of two between the scores on the dice is the winner of the game. Find the probability that
(i) A wins at his first throw [2]
The probability that A wins on his first throw is exactly the event described by (ii) above, and so the probability is 3/16.
[2 marks]
(ii) B wins at his first throw [5]
This event requires that A does not win on his first throw with probability 1 - 3/16 = 13/16, [1]
and B does win on his first throw with probability 3/16. [1]
Since these two events are independent, the probability of both occurring is the product of these probabilities, [2]
giving an overall probability of (13/16) X (3/16) = 39/256. [1]
[5 marks]
(ii) A wins at his third throw [5]
This event is equivalent to the following:
A does not win on his first throw (Prob = 13/16), B does not win on his first throw (Prob = 13/16), A does not win on his second throw (Prob = 13/16), B does not win on his second throw (Prob = 13/16), A wins on his third throw (Prob = 3/16). [3]
Each of these events is independent so the probability of all 5 happening is the product of the probabilities of each individual event, [1]
giving an overall probability of (13/16)4 X 3/16 = 0.082. [1]
[5 marks]