A pair of dice is thrown once. Find the probability that neither a doublet nor a total of 10 will appear.

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`