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(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 1/11. Do you agree with this argument? Justify your Solution:.

(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 1/11. Do you agree with this argument? Justify your Solution:.

Grade:12th pass

1 Answers

Pawan Prajapati
askIITians Faculty 60787 Points
2 years ago
If 2 dices are thrown, the possible events are: (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) So, the total numbers of events: 6×6 = 36 (i) It is given that to get the sum as 2, the probability is 1/36 as the only possible outcomes = (1,1) For getting the sum as 3, the possible events (or outcomes) = E (sum 3) = (1,2) and (2,1) So, P(sum 3) = 2/36 Similarly, E (sum 4) = (1,3), (3,1), and (2,2) So, P (sum 4) = 3/36 E (sum 5) = (1,4), (4,1), (2,3), and (3,2) So, P (sum 5) = 4/36 E (sum 6) = (1,5), (5,1), (2,4), (4,2), and (3,3) So, P (sum 6) = 5/36 E (sum 7) = (1,6), (6,1), (5,2), (2,5), (4,3), and (3,4) So, P (sum 7) = 6/36 E (sum 8) = (2,6), (6,2), (3,5), (5,3), and (4,4) So, P (sum 8) = 5/36 E (sum 9) = (3,6), (6,3), (4,5), and (5,4) So, P (sum 9) = 4/36 E (sum 10) = (4,6), (6,4), and (5,5) So, P (sum 10) = 3/36 E (sum 11) = (5,6), and (6,5) So, P (sum 11) = 2/36 E (sum 12) = (6,6) So, P (sum 12) = 1/36 So, the table will be as: (ii) The argument is not correct as it is already justified in (i) that the number of all possible outcomes is 36 and not 11.

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