We know that in a single throw of two dice, the total number of possible outcomes is `(6 xx 6) = 36`.
Let S be the sample space. Then, n(S) = 36.
(i) Let `E_(1)` = event of getting a doublet. Then,
`E_(1) = {(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6)}.`
`therefore n(E_(1)) 6.`
`therefore P(E_(1)) = (n(E_(1)))/(n(S)) = 6/36 = 1/6.`
(ii) Let `E_(2)` = event of getting an even number as the sum. Then,
`E_(2) = {(1, 1), (1, 3), (1, 5), (2, 2), (2, 4), (2, 6), (3, 3), (3, 1), (3, 5), (4, 2), (4, 4), (4, 6), (5, 1), (5, 3), (5, 5), (6, 2), (6, 4), (6, 6)}`.
`therefore n(E_(2)) = 18`.
`therefore P(E_(2)) = (n(E_(2)))/(n(S)) = 18/36 = 1/2.`
(iii) Let `E_(3)` = event of getting a prime number as the sum. Then,
`E_(3) = {(1, 1), (1, 2), (1, 4), (1, 6), (2, 1), (2, 3), (2, 5), (3, 2), (3, 4), (4, 1), (4, 3), (5, 2), (5, 6), (6, 1), (6, 5)}.`
`therefore n(E_(3)) = 15.`
`therefore P(E_(2)) = (n(E_(3)))/(n(S)) = 15/36 = 5/12.`
(iv) Let `E_(4)` = event of getting a multiple of 3 as the sum. Then,
`E_(4) = {(1, 2), (1, 5), (2, 1), (2, 4), (3, 3), (3, 6), (4, 2), (4, 5), (5, 1), (5, 4), (6, 3), (6, 6)}.`
`therefore n(E_(4)) = 12`.
`therefore P(E_(4)) = (n(E_(4)))/(n(S)) = 12/36 = 1/3.`
(v) Let `E_(5) =` event of getting a total of at least 10. Then,
`E_(5) =` event of getting a total of 10 or 11 or 12
`= {(4, 6), (5, 5), (5, 6), (6, 4), (6, 5), (6, 6)}.`
`therefore n(E_(5)) = 6.`
`therefore P(E_(5)) = (n(E_(5)))/(n(S)) = 6/36 = 1/6.`
(vi) Let `E_(6) =` event of getting a doublet of even numbers. Then,
`E_(6) = {(2, 2), (4, 4), (6, 6)}.`
`therefore n(E_(6)) = 3.`
`therefore P(E_(6)) = (n(E_(6)))/(n(S)) = 3/36 = 1/12.`
(vii) Let `E_(7) =` event of getting a multiple of 2 on one die and a multiple of 3 on the other die. Then,
`E_(7) = {(2, 3), (2, 6), (4, 3), (4, 6), (6, 3), (6, 6), (3, 2), (3, 4), (3, 6), (6, 2), (6, 4)}.`
`therefore n(E_(7)) = 11.`
`therefore P(E_(7)) = (n(E_(7)))/(n(S)) = 11/36.`
