Question

Two dice are numbered 1, 2, 3, 4, 5, 6 and 1, 1, 2, 2, 3, 3, respectively. They are thrown and the sum of the numbers on them is noted. Find the probability of getting each sum from 2 to 9 separately.

Answer 1

Number of total outcomes = 36

(i) Let E₁ = Event of getting sum 2 = {(1, 1), (1, 1)}

∴ n(E₁) = 2

∴ `P(E_1) = (n(E_1))/(n(S)) = 2/36 = 1/8`

(ii) Let E₂ = Event of getting sum 3 = {(1, 2), (1, 2), (2, 1), (2, 1)}

∴ n(E₂) = 4

∴ `P(E_2) = (n(E_2))/(n(S)) = 4/36 = 1/9`

(iii) Let E₃ = Event of getting sum 4 = {(2, 2), (2, 2), (3, 1), (3, 1), (1, 3), (1, 3)}

∴ n(E₃) = 6

∴ `P(E_3) = (n(E_3))/(n(S)) = 6/36 = 1/6`

(iv) Let E₄ = Event of getting sum 5 = {(2, 3), (2, 3), (4, 1), (4, 1), (3, 2), (3, 2)}

∴ n(E₄) = 6

∴ `P(E_4) = (n(E_4))/(n(S)) = 6/36 = 1/6`

(v) Let E₅ = Event of getting sum 6 = {(3, 3), (3, 3), (4, 2), (4, 2), (5, 1), (5, 1)}

∴ n(E₅) = 6

∴ `P(E_5) = (n(E_5))/(n(S)) = 6/36 = 1/6`

(vi) Let E₆ = Event of getting sum 7 = {(4, 3), (4, 3), (5, 2), (5, 2), (6, 1), (6, 1)}

∴ n(E₆) = 6

∴ `P(E_6) = (n(E_6))/(n(S)) = 6/36 = 1/6`

(vii) Let E₇ = Event of getting sum 8 = {(5, 3), (5, 3), (6, 2), (6, 2)}

∴ n(E₇) = 4

∴ `P(E_7) = (n(E_7))/(n(S)) = 4/36 = 1/9`

(viii) Let E₈ = Event of getting sum 9 = {(6, 3), (6, 3)}

∴ n(E₈) = 2

∴ `P(E_8) = (n(E_8))/(n(S)) = 2/36 = 1/18`