Given: a pair of dice has been thrown, so the number of elementary events in sample space is 62 = 36
n (S) = 36
By using the formula,
P (E) = favourable outcomes / total possible outcomes
(i) Let E be the event that the sum 8 appears
E = {(2, 6) (3, 5) (4, 4) (5, 3) (6, 2)}
n (E) = 5
P (E) = n (E) / n (S)
= 5 / 36
(ii) Let E be the event of getting a doublet
E = {(1, 1) (2, 2) (3, 3) (4, 4) (5, 5) (6, 6)}
n (E) = 6
P (E) = n (E) / n (S)
= 6 / 36
= 1/6
(iii) Let E be the event of getting a doublet of prime numbers
E = {((2, 2) (3, 3) (5, 5)}
n (E) = 3
P (E) = n (E) / n (S)
= 3 / 36
= 1/12
(iv) Let E be the event of getting a doublet of odd numbers
E = {(1, 1) (3, 3) (5, 5)}
n (E) = 3
P (E) = n (E) / n (S)
= 3 / 36
= 1/12
(v) Let E be the event of getting sum greater than 9
E = {(4,6) (5,5) (5,6) (6,4) (6,5) (6,6)}
n (E) = 6
P (E) = n (E) / n (S)
= 6 / 36
= 1/6
(vi) Let E be the event of getting even on first die
E = {(2,1) (2,2) (2,3) (2,4) (2,5) (2,6) (4,1) (4,2) (4,3) (4,4) (4,5) (4,6) (6,1) (6,2) (6,3) (6,4) (6,5) (6,6)}
n (E) = 18
P (E) = n (E) / n (S)
= 18 / 36
= 1/2
(vii) Let E be the event of getting even on one and multiple of three on other
E = {(2,3) (2,6) (4,3) (4,6) (6,3) (6,6) (3,2) (3,4) (3,6) (6,2) (6,4)}
n (E) = 11
P (E) = n (E) / n (S)
= 11 / 36
(viii) Let E be the event of getting neither 9 or 11 as the sum
E = {(3,6) (4,5) (5,4) (5,6) (6,3) (6,5)}
n (E) = 6
P (E) = n (E) / n (S)
= 6 / 36
= 1/6
(ix) Let E be the event of getting sum less than 6
E = {(1,1) (1,2) (1,3) (1,4) (2,1) (2,2) (2,3) (3,1) (3,2) (4,1)}
n (E) = 10
P (E) = n (E) / n (S)
= 10 / 36
= 5/18
(x) Let E be the event of getting sum less than 7
E = {(1,1) (1,2) (1,3) (1,4) (1,5) (2,1) (2,2) (2,3) (2,4) (3,1) (3,2) (3,3) (4,1) (4,2) (5,1)}
n (E) = 15
P (E) = n (E) / n (S)
= 15 / 36
= 5/12
(xi) Let E be the event of getting more than 7
E = {(2,6) (3,5) (3,6) (4,4) (4,5) (4,6) (5,3) (5,4) (5,5) (5,6) (6,2) (6,3) (6,4) (6,5) (6,6)}
n (E) = 15
P (E) = n (E) / n (S)
= 15 / 36
= 5/12
(xii) Let E be the event of getting neither a doublet nor a total of 10
E′ be the event that either a double or a sum of ten appears
E′ = {(1,1) (2,2) (3,3) (4,6) (5,5) (6,4) (6,6) (4,4)}
n (E′) = 8
P (E′) = n (E′) / n (S)
= 8 / 36
= 2/9
So, P (E) = 1 – P (E′)
= 1 – 2/9
= 7/9
(xiii) Let E be the event of getting odd number on first and 6 on second
E = {(1,6) (5,6) (3,6)}
n (E) = 3
P (E) = n (E) / n (S)
= 3 / 36
= 1/12
(xiv) Let E be the event of getting greater than 4 on each die
E = {(5,5) (5,6) (6,5) (6,6)}
n (E) = 4
P (E) = n (E) / n (S)
= 4 / 36
= 1/9
(xv) Let E be the event of getting total of 9 or 11
E = {(3,6) (4,5) (5,4) (5,6) (6,3) (6,5)}
n (E) = 6
P (E) = n (E) / n (S)
= 6 / 36
= 1/6
(xvi) Let E be the event of getting total greater than 8
E = {(3,6) (4,5) (4,6) (5,4) (5,5) (5,6) (6,3) (6,4) (6,5) (6,6)}
n (E) = 10
P (E) = n (E) / n (S)
= 10 / 36
= 5/18
