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 on them is 5 with probability

• A. $\frac{1}{8}$
• B. $\frac{13}{15}$
• C. $\frac{1}{7}$
• D. None of these

Answer 1

The correct option is B $\frac{13}{15}$

Let A be the event that the maximum number on the two chosen tickets is not more than 10 i.e., the number on them $\le 10$ and B be the event that the maximum number on them is 5, i.e., the number on them is $\ge 5$ we have to find $P\left(\frac{B}{A}\right)$.
Now $\left(\frac{B}{A}\right)=\frac{P(A\cap B)}{P\left(A\right)}=\frac{n(A\cap B)}{n\left(A\right)}$
Now the number of ways of getting a number r on the two tickets is the coefficient of x'in the expansion of
$(x^2+x^2+x^3+\dots +x^{100}{)}^2=x^2(1+x+\dots +x^{99}{)}^2$
$=x^2{\left(\frac{1-x^{100}}{1-x}\right)}^2=x^2(1-2x^{100}+x^{200})(1-x{)}^{-2}$
$=x^2(1-2x^{100}+x^{200})(1+2x+3x^2+\dots +(r+1)x^4+\dots )$ Thus coefficient of $x^2=1$, of $x^3=2$, of $x^4=3\dots of{x}^{10}$ 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 $x^2,x^3,\dots \dots x^{10}$ and in $n(A\cap B)$ we add the coefficient of $x^5,x^6,\dots \dots x^{10}$ ]
Hence required probability =$\frac{39}{45}=\frac{13}{15}$.