A box contains 100 tickets, numbered 1, 2, ....100. Two tickets are chosen at random one after another with replacement. It is given that the maximum sum on the two chosen tickets is not more than 10. The minimum sum of them is 5 with probability 

For getting sum-2possi$\to 1$

                              3possi$\to 2$

                               .

                               .

                               .          

                               10possi$\to 9$

$P\left(E_1\right)=45$

$P\left(E_1\cap E_2\right)=39$

$P\left(\frac{E_2}{E_1}\right)=\frac{39}{45}=\frac{13}{15}$