Complete the following table : Event : 'Sum on two dice' | Probability -------- | --------- 2 | 1/36 3 | 4 | 5 | 6 | 7 | 8 | 5/36 9 | 10 | 11 | 12 | 1/36 (ii) A student argues that there are 11 possible outcomes 2, 3, 4, 5, 6, 7, 8, 9, 10, 11 and 12. Therefore, each of them has a probability . Do you agree with this argument? Justify your answer.

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KnowledgeBoat · Accepted Answer

On tossing two dice. Sample space = {(1,1) (1,2) (1,3) (1,4) (1,5) (1,6) (2,1) (2,2) (2,3) (2,4) (2,5) (2,6) (3,1) (3,2) (3,3) (3,4) (3,5) (3,6) (4,1) (4,2) (4,3) (4,4) (4,5) (4,6) (5,1) (5,2) (5,3) (5,4) (5,5) (5,6) (6,1) (6,2) (6,3) (6,4) (6,5) (6,6)}. No. of possible outcomes = 36 Favourable outcomes for getting sum of 3 are {(1, 2), (2, 1)} No. of favourable outcomes = 2 P(getting a sum of 3) = . Favourable outcomes for getting sum of 4 are {(1, 3), (2, 2), (3, 1)} No. of favourable outcomes = 3 P(getting a sum of 4) = . Favourable outcomes for getting sum of 5 are {(1, 4), (2, 3), (3, 2), (4, 1)} No. of favourable outcomes = 4 P(getting a sum of 5) = . Favourable outcomes for getting sum of 6 are {(1, 5), (2, 4), (3, 3), (4, 2), (5, 1)} No. of favourable outcomes = 5 P(getting a sum of 6) = . Favourable outcomes for getting sum of 7 are {(1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1)} No. of favourable outcomes = 6 P(getting a sum of 7) = . Favourable outcomes for getting sum of 9 are {(3, 6), (4, 5), (5, 4), (6, 3)} No. of favourable outcomes = 4 P(getting a sum of 9) = . Favourable outcomes for getting sum of 10 are {(4, 6), (5, 5), (6, 4)} No. of favourable outcomes = 3 P(getting a sum of 10) = . Favourable outcomes for getting sum of 11 are {(5, 6), (6, 5)} No. of favourable outcomes = 2 P(getting a sum of 11) = . Event : 'Sum on two dice' | Probability -------- | --------- 2 | 1/36 3 | 2/36 4 | 3/36 5 | 4/36 6 | 5/36 7 | 6/36 8 | 5/36 9 | 4/36 10 | 3/36 11 | 2/36 12 | 1/36 (ii) No, I don't agree because the argument is false. The eleven sums are not equally likely. Hence, each of them cannot have a probability of . Also, in the above table we see that each sum has different probability. Hence, the above argument is false.

Mathematics

Probability

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