Let S be the sample space. ∴ n ( S ) = 36 Let E = event of getting a doublet = { ( 1 , 1 ) , ( 2 , 2 ) , ( 3 , 3 ) , ( 4 , 4 ) , ( 5 , 5 ) , ( 6 , 6 ) } ⇒ n ( E ) = 6 Let F = event of getting a sum of 10 = { ( 4 , 6 ) , ( 5 , 5 ) , ( 6 , 4 ) } ⇒ n ( f ) = 3 Now E ∩ F = { ( 5 , 5 ) } ⇒ n ( E ∩ F ) = 1 ∴ P ( E ∪ F ) = P ( E ) + P ( F ) − P ( E ∩ F ) − 6 36 + 3 36 − 1 36 = 8 36 = 2 9 Now the probability that neither a doublet nor a sum of 10 will appear = P ( ¯¯¯ E ∩ ¯¯¯ F ) = P ( ¯¯¯¯¯¯¯¯¯¯¯ E ∪ F ) = 1 − P ( E ∪ F ) = 1 − 1 9 = 7 9

Answer

N/a

Explanation

Let `S` be the sample space.
`:.n(S)=36`
Let `E=` event of getting a doublet
`={(1,1),(2,2),(3,3),(4,4),(5,5),(6,6)}`
`implies n(E)=6`
Let `F=` event of getting a sum of 10
`={(4,5),(5,5),(6,4)}`
`implies n(f)=3`
Now `EnnF={(5,5)}`
`implies n(EnnF)=1`
`:. P(EuuF)=P(E)+P(F)-P(EnnF)`
`-6/36+3/36-1/36`
`=8/36=2/9`
Now the probability that neither a doublet nor a sum of 10 will appear
`=P(barEnnbarF)`
`=P(bar(EuuF))`
`=1-P(EuuF)`
`=1-1/9=7/9`