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

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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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$$P(n)=Kn2Given$$      $$P(1)=K$$,$$P(2)=22K$$,$$P(3)=32K$$,$$P(4)=42K$$,$$P(5)=52K$$,$$P(6)=62KSo$$, 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, $$P(A)$$ $$=$$ $$P(C)Also$$, $$P(A)$$ $$+$$ $$P(B)$$ $$+$$ $$P(C)$$ $$=$$ $$1Since$$   $$P(B)=$$∑$$i=16$$ [$$P(i)$$]$$2$$                 $$=1+16+81+256+625+129691$$×$$91=227591$$×$$91=2591Now$$, $$2P(A)$$ $$+$$ $$P(B)$$ $$=$$ $$1$$⇒ $$P(A)=12$$[$$1$$−$$P(B)$$]$$=3391$$

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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