A die is weighted such that the probability of rolling the face numbered n is proportional to n2(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=12⎛⎜⎜⎜⎜⎜⎝1−x∑n=1n4y2⎞⎟⎟⎟⎟⎟⎠, then the value of x+y is(where [.] represents the greatest integer function)

A die is weighted such that the probability of rolling the face number

1.	A die is weighted such that the probability of rolling the face numbered n is proportional to n2(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=12⎛⎜⎜⎜⎜⎜⎝1−x∑n=1n4y2⎞⎟⎟⎟⎟⎟⎠, then the value of x+y is(where [.] represents the greatest integer function)
Answer» 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 $P = \frac{1}{2} ⎛ ⎜⎜⎜⎜⎜ ⎝ 1 - \frac{x \sum n = 1 n^{4}}{y^{2}} ⎞ ⎟⎟⎟⎟⎟ ⎠,$ then the value of $x + y$ is (where [.] represents the greatest integer function)