Question

A die is weighted such that the probability of rolling the face numbered $n$ is proportional to $n^2(n=1,2,3,4,5,6)$. The die is rolled twice, yielding the number $a$ and $b$. The probability that $a<b$ is $P$ such that $P=\frac{1}{2}(1-\frac{\sum _{n=1}^x{n}^4}{y^2}),$ then the value of $x+y$ is
(where [.] represents the greatest integer function)

Answer 1

$P\left(n\right)\propto n^2\Rightarrow P\left(n\right)=\lambda n^2$ where $\sum _{n=1}^{6}P\left(n\right)=1\Rightarrow \lambda \left(\sum _{n=1}^6{n}^2\right)=1\Rightarrow \lambda \left[\frac{6\times 7\times 13}{6}\right]=1\Rightarrow \lambda =\frac{1}{91}$
We know that $P(a<b)$ and $P(a>b)$ are equally likely cases. So, $P(a<b)=P(a>b)\cdots \left(i\right)$
And the sum of all mutually exclusive events, $P(a<b)+P(a=b)+P(a>b)=1$
$P(a<b)=\frac{1}{2}[1-P(a=b\left)\right]\left(\text{From}\right(i\left)\right)$
$\Rightarrow P=\frac{1}{2}(1-\frac{1^4+2^4+...+6^4}{91\times 91})$
$\therefore x=6,y=91$
$\Rightarrow x+y=97$