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A dice is weighted such that the probabi...

A dice 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 dice is rolled twice, yielding the number a and b. The probability that `a gt b` is p then the value of `[2//p]` (where [ . ] represents greatest integer function) is _______.

Text Solution

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`P(n) = Kn^(2)`
Given `P(1) = K, P(2) = 2^(2)K, P(3) = 3^(2)K, P(4)= 4^(2)K`,
`P(5) = 5^(2)K, P(6) = 6^(2)K`
So, total probability = 91K
implies 1 = 91K
`therefore K = (1)/(91)`
Therefore, P(1) = `(1)/(91)`, `P(2) = (4)/(91)` and so on.
Let three events A, B and C be defined as
`A: a lt b`
B: a = b
`C: a gt b`
By symmetry, P(A) = P(C )
Also, P(A) + P(B) + P(C ) = 1
Since P(B) = `=underset(i = 1) overset(6)sum[P(i)]^(2) = [(1+16+81+256+625+1296)/(91xx91)]`
`=(2275)/(91xx91)=(25)/(91)`
Now, 2P(A) + P(B) = 1
`implies P(A) = (1)/(2) [1 - P(B)]=(33)/(91)`
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