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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 _______.

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Detailed Solution

P(n)=Kn2Given P(1)=K,P(2)=22K,P(3)=32K,P(4)=42K,P(5)=52K,P(6)=62KSo, total probability = 91 K⇒ 1=91K∴ K=191Therefore, P(1)=191;P(2)=491 and so on.Let three events A, B and C be defined asA:abBy symmetry, P(A) = P(C)Also, P(A) + P(B) + P(C) = 1Since P(B)=∑i=16 [P(i)]2 =1+16+81+256+625+129691×91=227591×91=2591Now, 2P(A) + P(B) = 1⇒ P(A)=12[1−P(B)]=3391

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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 _______.