Each of 2 urns A and B has n balls 1 to n. One ball is drawn from each of the urns. Probability that ball drawn from urn A bears a smaller number than the ball drawn from urn B, is 99/200. Value of n is?    [100]

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$$\begin{gathered} \mathrm{It} \mathrm{is} \mathrm{equally} \mathrm{likely} \mathrm{that} \mathrm{the} \mathrm{ball} \mathrm{drawn} \mathrm{from} \mathrm{urn} \mathrm{A} \mathrm{will} \mathrm{bear} \mathrm{a} \mathrm{number} \\ \mathrm{larger} \mathrm{or} \mathrm{smaller} \mathrm{than} \mathrm{that} \mathrm{of} \mathrm{B} \mathrm{and} \mathrm{the} \mathrm{probability} \mathrm{of} \mathrm{eqal} \mathrm{numbers} \mathrm{is} \\ \frac{1}{\mathrm{n}}. \mathrm{Hence} \\ \mathrm{p}+\mathrm{p}+\frac{1}{\mathrm{n}}=1 \\ \Rightarrow 2\mathrm{p}=1-\frac{1}{\mathrm{n}} \\ \Rightarrow \mathrm{p}=\frac{\mathrm{n}-1}{2\mathrm{n}} \\ \mathrm{According} \mathrm{to} \mathrm{question} \\ \frac{\mathrm{n}-1}{2\mathrm{n}}=\frac{99}{200} \\ \Rightarrow 200\mathrm{n}-200=198\mathrm{n} \\ \Rightarrow 2\mathrm{n}=200 \\ \mathbf{\Rightarrow }\mathbf{n}\mathbf{=}\mathbf{100}\mathbf{ }\left(\mathbf{Answer}\right) \end{gathered}$$

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