Question

Two dice are numbered 1,2,3,4,5,6 and 1,1,2,2,3,3, resepectively. They are thrown and the sum of the number is noted. Find the probability of getting each sum from 2 to 9, seperately.

Answer 1

Number of total outcomes =n(s)=36
(i) Let, `E_(1)` =Event of getting sum 2={(1,1),(1,)}
`n(E_(1))=2`
`P(E_(1))=(n(E_(1)))/(n(S))=(2)/(36)=(1)/(18)`
(ii) Let, `E_(2)`=Event of getting sum 3= {(1,2),(1,2),(2,1),(2,1)}
`n(E_(2))=4`
`P(E_(2))=(n(E_(2)))/(n(S))=(4)/(36)=(1)/(9)`
Let `E_(3)`=Event of getting sum 4={(2,2),(2,2),(3,1),(3,1),(1,3),(1,3)}
`n(E_(3)=4`
`P(E_(3))=(n(E_(3)))/(n(S))=(6)/(36)=(1)/(6)`
(iv) Let `E_(4)`=Event of getting sum `5={(2,3),(2,3),(4,1),(4,1),(3,2),(3,2)}`
`n(E_(4))=6`
`P(E_(4))=(n(E_(4)))/(n(S))=(6)/(36)=(1)/(6)`
(v) Let `E_(5)`=Event of getting sum 6 ={(3,3),(3,3),(4,2)(4,2),(5,1),(5,1)}
`n(E_(5))=6`
`P(E_(5))=(n(E_(5)))/(n(S))=(6)/(36)=(1)/(6)`
(vi) Let `E_(6)`=Event of getting sum 7={(4,3),(4,3),(5,2),(5,2),(6,1),(6,1)}
`n(E_(6))=6`
`P(E_(6))=(n(E_(6)))/(n(S))=(6)/(36)=(1)/(6)`
(vii) Let `E_(7)`= Event of getting sum 8={(5,3),(5,3),(6,2),(6,2)}
`n(E_(7))=4`
`P(E_(7))=(n(E_(7)))/(n(S))=(4)/(36)=(1)/(9)`
(viii) Let `E_(8)`=Event of getting sum 9={(6,3),(6,3)}
`n(E_(8))=2`
`P(E_(8))=(n(E_(8)))/(n(S))=(2)/(36)=(1)/(18)`