Two dice, one blue and one grey, are thrown at the same time.
Two dice, one blue and one grey, are thrown at the same time.
(i) Write down all the possible outcomes and complete the following table:
(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 $\frac{1}{11}$. Do you agree with this argument?
(i) It can be observed that,
To get the sum as 2, possible outcomes = (1, 1)
To get the sum as 3, possible outcomes = (2, 1) and (1, 2)
To get the sum as 4, possible outcomes = (3, 1), (1, 3), (2, 2)
To get the sum as 5, possible outcomes = (4, 1), (1, 4), (2, 3), (3, 2)
To get the sum as 6, possible outcomes = (5, 1), (1, 5), (2, 4), (4, 2),
(3, 3)
To get the sum as 7, possible outcomes = (6, 1), (1, 6), (2, 5), (5, 2),
(3, 4), (4, 3)
To get the sum as 8, possible outcomes = (6, 2), (2, 6), (3, 5), (5, 3),
(4, 4)
To get the sum as 9, possible outcomes = (3, 6), (6, 3), (4, 5), (5, 4)
To get the sum as 10, possible outcomes = (4, 6), (6, 4), (5, 5)
To get the sum as 11, possible outcomes = (5, 6), (6, 5)
To get the sum as 12, possible outcomes = (6, 6)
(ii)Probability of each of these sums will not be $\frac{1}{11}$ as these sums are not equally likely.