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

Let A be the event that the maximum number on the two chosen tickets is not more than 10, that is, the no. on them $\le 10$ and B the event that the minimum no. on them is 5, that is, the no. on them is $\ge 5.$We have to find $P(B/A)$

Now $P(B/A)=\frac{P(A\cap B)}{P\left(A\right)}=\frac{n(A\cap B)}{n\left(A\right)}$

Now the number of ways of getting a number r on the two tickets is the coeff. of $x^7$in the expansion of 

${\left(x^1+x^2+\dots +x^{100}\right)}^2=x^2{\left(1+x+\dots +x^{99}\right)}^2$

$=x^2{\left(\frac{1-x^{100}}{1-x}\right)}^2=x^2\left(1-2x^{100}+x^{200}\right)(1-x{)}^{-2}$

$=x^2\left(1-2x^{100}+x^{200}\right)\left(1+2x+3x^2+\dots (r+1)x^r+\dots \right)$

Thus coeff. of $x^2=1\text{, coeff. of }{x}^3=2\text{, coeff. of }{x}^4=3\text{, }$$\dots ,\text{ coeff. of }{x}^{10}$is 9.

Hence, $n\left(A\right)=1+2+3+4+5+6+7+8+9=45$

and $n(A\cap B)=4+5+6+7+8+9=39$

[Note that in finding n(A), we have to add the coefficients of $x^2,x^3,\dots ,x^{10}$are in $n(A\cap B)$ we add the coefficients of $\left.x^5,x^6,\dots ,x^{10}\right]$

Hence, required probability $=P(B/A)=\frac{39}{45}=\frac{13}{15}$