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Nov 2002 p6 q2
2575
Ivan throws three fair dice.
List all the possible scores on the three dice which give a total score of 5, and hence show that the probability of Ivan obtaining a total score of 5 is \(\frac{1}{36}\).
Find the probability of Ivan obtaining a total score of 7.
Solution
(i) To find the probability of obtaining a total score of 5, we list all possible combinations of scores on the three dice that sum to 5:
(1, 1, 3)
(1, 2, 2)
(1, 3, 1)
(2, 1, 2)
(2, 2, 1)
(3, 1, 1)
There are 6 combinations. Since each die has 6 faces, the total number of possible outcomes when throwing three dice is \(6^3 = 216\). Thus, the probability is \(\frac{6}{216} = \frac{1}{36}\).
(ii) To find the probability of obtaining a total score of 7, we list all possible combinations of scores on the three dice that sum to 7:
(1, 2, 4) and permutations: 6 combinations
(1, 3, 3) and permutations: 3 combinations
(2, 2, 3) and permutations: 3 combinations
(1, 1, 5) and permutations: 3 combinations
There are 15 combinations. Thus, the probability is \(\frac{15}{216} = \frac{5}{72}\).
