A dice is weighted such that the probability of rolling the face numbered n is proportional to n² (n = 1, 2, 3, 4, 5, 6). The dice is rolled twice, yielding the numbers a and b. The probability that a < b is p then the value of [2/p] (where [ . ] represents greatest integer function) is _______.

$\mathrm{P}\left(\mathrm{n}\right)={\mathrm{Kn}}^2$

$\begin{array}{c}\mathrm{Given} \mathrm{P}\left(1\right)=\mathrm{K},\mathrm{P}\left(2\right)=2^2\mathrm{K},\mathrm{P}\left(3\right)=3^2\mathrm{K},\mathrm{P}\left(4\right)=4^2\mathrm{K},\\ \mathrm{P}\left(5\right)=5^2\mathrm{K},\mathrm{P}\left(6\right)=6^2\mathrm{K}\end{array}$

So, total probability = 91 K

$\Rightarrow 1=91\mathrm{K}$

$\therefore \mathrm{K}=\frac{1}{91}$

Therefore, $\mathrm{P}\left(1\right)=\frac{1}{91};\mathrm{P}\left(2\right)=\frac{4}{91}$ and so on.

Let three events A, B and C be defined as

$\begin{array}{r}\mathrm{A}:\mathrm{a}<\mathrm{b}\\ \mathrm{B}:\mathrm{a}=\mathrm{b}\\ \mathrm{C}:\mathrm{a}>\mathrm{b}\end{array}$

By symmetry, P(A) = P(C)

Also, P(A) + P(B) + P(C) = 1

Since   $\mathrm{P}\left(\mathrm{B}\right)=\sum _{\mathrm{i}=1}^6 \left[\mathrm{P}\right(\mathrm{i}){]}^2$

                 $\begin{array}{l}=\left[\frac{1+16+81+256+625+1296}{91\times 91}\right]\\ =\frac{2275}{91\times 91}=\frac{25}{91}\end{array}$

Now, 2P(A) + P(B) = 1

$\Rightarrow \mathrm{P}\left(\mathrm{A}\right)=\frac{1}{2}[1-\mathrm{P}(\mathrm{B}\left)\right]=\frac{33}{91}$

$\frac{2}{p}=\frac{2\times 91}{33}=\frac{182}{33}\Rightarrow \left[\frac{182}{33}\right]=5$