December
1998 QUESTION 2 Total Marks: 20 Marks |
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questions
SUGGESTED SOLUTIONS
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| (a) | (i) In how many ways can a set of five different letters be selected from the English alphabet? | [1] |
| 26C5 =
65,780
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| (ii) How many subsets with at least two elements does a set with six elements have? | [3] | |
| 1 mark
for calculating the the total number of subsets that a set with six elements has is 26 = 64. 1 mark for the correct answer i.e., 64 - 7 = 57 subsets.
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| (iii) A group contains n men and n women. How many ways are there to arrange these people in a row if the men and women alternate? | [3] | |
| 1 mark for
observing that the first person can be either a man or a woman and so can be any one of
the 2n people. 1 mark for observing that the gender of the second person (and indeed all the rest) is determined by the gender of the first person in the row and so the second person can be one from a choice of n. The third and fourth can each be one from a choice of n - 1 and so on. 1 mark for calculating the correct probability of 2 * n! This is not the same as (2n)!
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| (b) | A tetrahedral die is one with four sides. The sides are numbered 1,2,3 and 4. When a tetrahedral die is thrown, the score is the number on the face which ends up face down. | |
| (i) Write down the sample space for the experiement consisting of throwing 2 tetrahedral dice. | [2] | |
| {(1,1),
(1,2), (1,3), (1,4), (1,1), (1,2), (1,3), (1,4), (1,1), (1,2), (1,3), (1,4), (1,1), (1,2), (1,3), (1,4)}
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| (ii) Write down the subset of the sample space corresponding to the event that the sum of the scores is 5. | [2] | |
| {(1,4),
(2,3), (3,2), (4,1)} Subtract one mark for each error or omission (up to a max of two).
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| (iii) Calculate the possibility of the event that the sum of the result is 5. | [2] | |
| 4/16 =
1/4 Two marks for the correct answer (not necessarily in its simplified form); otherwise award one mark if either the numerator or the denominator is correct.
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| (c) | A restaurant is doing a special "Dinner for Two Meal Deal". Customers choose any two different main courses from a selection of eight; any three different side dishes from a selection of ten and any two different deserts from a selection of six. | |
| (i) How many possible means can be chosen? | [4] | |
| 8C2
x 10C3 x 6C2 = 28 x 120 x 15 = 50,400 1 mark for 8C2; 1 mark for 10C3; 1 mark for 6C2; 1 mark for the answer.
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| (ii) Five of the main courses and one of the side dishes contain meat. If all possible combinations are equally likely to be chosen, what is the probability that the meal does not contain meat. | [3] | |
| 1 mark
for calculating the number of possible vegetarian meals i.e., 3C2 x 9C3 x 6C2 = 3 x 84 x 15 = 3780 1 mark for the dividing this answer by the answer obtained in the previous part. 1 mark for the correct answer i.e., 3780/50400 = 3/40 = 0.075
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