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. What is the probability that the minimum number on them is not less than 5?

• 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 minimum number on them is not less than 5, i.e., the number on them is $\ge 5$ we have to find $P\left(\frac{B}{A}\right)$.

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

Now the number of ways of getting a number $r$ as the maximum on the two tickets is the coefficient of $x^r$ in the expansion of

$(x^1+x^2+x^3+\dots \dots +x^{100})(x^1+x^2+x^3+\dots \dots +x^{100})=x^2(1+x+\dots \dots +x^{99}{)}^2$

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

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

Thus coefficient of $x^2=1$ of $x^3=2$, of $x^4=3$ ..... of $x^{10}$ is 9.

Hence n(A) = 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 coefficients of $x^5,x^6,\dots \dots x^{10}$]

Hence required probability $=\frac{39}{45}=\frac{13}{15}$.

OR

Choosing 2 tickets out 10 (1 to 10) $={{}^{10}C}_2$

No of ways with 2 as maximum number = 1 [(1,2)]

No of ways with 3 as maximum number = 2 [(1,3) (2,3)]

No of ways with 4 as maximum number = 3 [(1,4) (2,4) (3,4)]

$\therefore$ required probability $=\frac{{{}^{10}C}_2-(1+2+3)}{{{}^{10}C}_2}$

$=\frac{45-6}{45}=\frac{13}{15}$