Question

# Question 23 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.

Solution

## Number of total outcomes = n(s) = 36 (i) Let E1 = Event of getting sum 2 = {(1,1), (1,1)} n(E1)=2 ∴P(E1)=n(E1)n(S)=236=118 (ii) Let E2 = Event of getting sum = 3 {(1,2), (1,2),(2,1), (2,1)} n(E2)=4 ∴P(E2)=n(E2)n(S)=436=19 (iii) Let E3 = Event of getting sum 4 = {(2,2), (2,2), (3,1), (3,1), (1,3), (1,3)} ∴n(E3)=6 ∴P(E3)=n(E3)n(S)=636=16. (iv) Let E4 = Event of getting sum 5 = {(2,2),(2,3)(4,1), (4,1), (3,2), (3,2)} ∴n(E4)=6 ∴P(E4)=n(E4)n(S)=636=16 (v) Let E5 = Event of getting sum 6 = {(3,3), (3,3), (4,2), (4,2), (5,1), (5,1)} n(E5)=6 ∴P(E5)=n(E5)n(S)=636=16 (vi) Let E6 = Event of getting sum 7 = {(4,3), (4,3), (5,2), (5,2), (6,1), (6,1)} ∴n(E6)=6 ∴P(E6)=n(E6)n(S)=636=16 (vii) Let E7 = Event of getting sum 8 = {(5,3), (5,3), (6,2), (6,2)} ∴n(E7)=4 ∴P(E7)=n(E7)n(S)=436=19 (viii) Let E8 = Event of getting sum 9 = {(6,3), (6,3)} ∴n(E8)=2 ∴P(E8)=n(E8)n(S)=236=118MathematicsNCERT ExemplarStandard X

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