Question

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 $(15-k)/15$. Find the value of $k.$

Answer 1

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 them is $\ge$ $5$.
We have to find $P\left(B/A\right)$.
Now $P\left(B/A\right)=\frac{P(A\cap B)}{P\left(A\right)}=\frac{n(A\cap B)}{n\left(A\right)}.$
Now the number of ways of getting a number $r$ on the two tickets is the
coeff. of $x^r$ in the expansion of ${(x^1+x^2+....+x^{100})}^2=x^2{(1+x+....+x^{99})}^2$
$x^2{\left(\frac{1-x^{100}}{1-x}\right)}^2=x^2(1-2x^{100}+x^{200}){(1-x)}^{-2}$
$x^2(1-2x^{100}+x^{200})(1+2x+3x^2+\cdots (r+1)x^r+\cdots )$
Thus coeff. of $x^2=1,of{x}^3=2,of{x}^4=3,\cdots ,of{x}^{10}$ 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 $x^2,x^3,\cdots ,x^{10}$ and in $n(A\cap B)$ we add the coefficients of $x^2,x^3,\cdots x^{10}$ ].
Hence required prob. $=P\left(B/A\right)=\frac{39}{45}=\frac{13}{15}.$