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In one throw of a pair of dice, the sample space n ( S ) = 36 (i) Let E 1 = event of getting the sum of numbers on two dice 9 = { ( 3 , 6 ) , ( 4 , 5 ) , ( 5 , 4 ) , ( 6 , 3 ) } ⇒ n ( E 1 ) = 4 and P ( E 1 ) = n ( E 1 ) n ( S ) = 4 36 = 1 9 (ii) Let E 2 = event of getting the sum of numbers on two dice at least 9 = { ( 3 , 6 ) , ( 4 , 5 ) , ( 5 , 4 ) , ( 6 , 3 ) , ( 4 , 6 ) , ( 5 , 5 ) , ( 6 , 4 ) , ( 5 , 6 ) , ( 6 , 5 ) , ( 6 , 6 ) } ⇒ n ( E 2 ) = 10 and P ( E 2 ) = n ( E 2 ) n ( S ) = 10 36 = 5 18

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In one throw of a pair of dice, the sample space
`n(S)=36`
(i) Let `E_(1)=` event of getting the sum of numbers on two dice 9
`={(3,6),(4,5),(5,4),(6,3)}`
`implies n(E_(1))=4`
and `P(E_(1))=(n(E_(1)))/(n(S))=4/36=1/9`
(ii) Let `E_(2)=` event of getting the sum of numbers on two dice at least 9
`={(3,6),(4,5),(5,4),(6,3),(4,6),(5,5),(6,4),(5,6),(6,5),(6,6)}`
`implies n(E_(2))=10`
and `P(E_(2))=(n(E_(2)))/(n(S))=10/36=5/18`
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