In a throw of pair of dice, total no of possible outcomes = 36 (6 × 6) which 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)
(i) Let E be event of getting the sum as 8
No. of favorable outcomes = 5 {(2, 6) , (3, 5) , (4, 4) , (5, 3) , (6, 2)}
We know that, Probability P(E) = (No.of favorable outcomes)/(Total no.of possible outcomes)
P(E) = 5/36
(ii) E ⟶ event of getting a doublet
No. of favorable outcomes = 5 {(1, 1) , (2, 2) , (3, 3) , (4, 4) , (5, 5) , (6, 6)}
Total no. of possible outcomes = 36
P(E) = 6/36 = 1/6
(iii) E ⟶ event of getting a doublet of prime no’s
No. of favorable outcomes = 3 {(2, 2) , (3, 3) , (5, 5)}
Total no. of possible outcomes = 36
P(E) = 3/36 = 1/12
(iv) E ⟶ event of getting a doublet of odd no’s
No. of favorable outcomes = 3 {(1, 1) , (3, 3) , (5, 5)}
Total no. of possible outcomes = 36
P(E) = 3/36 = 1/12
(v) E ⟶ event of getting a sum greater than 9
No. of favorable outcomes = 6 {(4, 6) , (5, 5) , (5, 6) , (6, 4) , (6, 5) , (6, 6)}
Total no. of possible outcomes = 36
P(E) = 6/36 = 1/6
(vi) E ⟶ event of getting an even no. on first
No. of favorable outcomes = 18 {(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)}
Total no. of possible outcomes = 36
P(E) = 18/36 = 1/2
(vii) E ⟶ event of getting an even no. on one and a multiple of 3 on other
No. of favorable outcomes = 11 {(2, 3) , (2, 6) , (4, 3) , (4, 6) , (6, 3) , (6, 6) , (3, 2) , (3, 4) , (3, 4) , (3, 6) , (6, 2) , (6, 4)}
Total no. of possible outcomes = 36
P(E) = 11/36
(viii) Bar E ⟶ event of getting neither 9 nor 11 as the sum of numbers on faces
E ⟶ getting either 9 or 11 as the sum of no’s on faces
No. of favorable outcomes = 6 {(3, 6) , (4, 5) , (5, 4) , (6, 3) , (5, 6) , (6, 5)}
Total no. of possible outcomes = 36
P(E) = 6/36 = 1/6
P(Bar E) = 1 - P = 1 - 1/6 = 5/6
(ix) E ⟶ event of getting a sum less than 6
No. of favorable outcomes = 10 {(1, 1) , (1, 2) , (1, 3) , (1, 4) , (2, 1) , (2, 2) , (2, 3) , (3, 1) , (3, 2) , (4, 1)}
Total no. of possible outcomes = 36
P(E) = 10/36 = 5/18
(x) E ⟶ event of getting a sum less than 7
No. of favorable outcomes = 15 {(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)}
Total no. of possible outcomes = 36
P(E) = 15/36 = 5/12
(xi) E ⟶ event of getting a sum more than 7
No. of favorable outcomes = 15 {(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)}
Total no. of possible outcomes = 36
P(E) = 15/36 = 5/12
(xii) E ⟶ event of getting a 1 at least once
No. of favorable outcomes = 11 {(1, 1) , (1, 2) , (1, 3) , (1, 4) , (1, 5) , (1, 6) , (2, 1) , (3, 1) , (4, 1) , (5, 1) , (6, 1)}
Total no. of possible outcomes = 36
P(E) = 11/36
(xiii) E ⟶ event of getting a no other than 5 on any dice
No. of favourable outcomes = 25 {(1, 1) , (1, 2) , (1, 3) , (1, 4) , (1, 6) , (2, 1) , (2, 2) , (2, 3) , (2, 4) , (2, 6) , (3, 1) , (3, 2) , (3, 3) , (3, 4) , (3, 6) , (4, 1) , (4, 2) , (4, 3) , (4, 4) , (4, 6) , (6, 1) , (6, 2) , (6, 3) , (6, 4) , (6, 6)}
Total no. of possible outcomes = 36
P(E) = 25/36