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

We know Probability P ( E ) = $\frac{Totalnumberofdesiredeventsn(E\hspace{0.17em})}{Totalnumberofeventsn(S\hspace{0.17em})}$

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 = $\frac{22}{100}=\frac{\mathbf{11}}{\mathbf{50}}\mathbf{}\mathbf{(}\mathbf{}\mathit{A}\mathit{n}\mathit{s}\mathbf{}\mathbf{)}$

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