What is the probability that a no.being picked from 1-100 is divisible by 3 but not by 9

Answer :

We know Probability P ( E ) = $\frac{Total number of desired events n ( E\hspace{0.17em})}{Total number of events n ( S\hspace{0.17em})}$
Here two dice are thrown simultaneously .
So,
Total number of events n ( S ) = 100

1 ) Getting a  number divisible by 3 but not by 9 ,  We get combination  = { 3 ,  6 ,  12 , 15 ,  21  , 24 ,  30 , 33 , 39 , 42 , 48 , 51 , 57 , 60 , 66 , 69 , 75 , 78 , 84 , 87 , 93 , 96 }

Total  number of combinations  =  22

n ( E ) = 22

So,
Probability of getting a number divisible by 3 but not by 9 = $\frac{22}{100} = \frac{\mathbf{11}}{\mathbf{50}} \mathbf{ }\mathbf{(}\mathbf{ }\mathit{A}\mathit{n}\mathit{s}\mathbf{ }\mathbf{)}$