A bag contains a large number of white and black marbles in equal proportions. Two samples of 5 marbles are selected (with replacement) at random. The probability that the first sample contains exactly 1 black marble, and the second sample contains exactly 3 black marbles, is

Let the number of marble be 2n (where n is large)

Required probability $=\underset{\mathrm{n}\to \infty }{\lim }\frac{\mathrm{n}\times {}^{\mathrm{n}}\mathrm{C}_4}{{}^{2\mathrm{n}}\mathrm{C}_5}\times \frac{{}^{\mathrm{n}}\mathrm{C}_3\times {}^{\mathrm{n}}\mathrm{C}_2}{{}^{2\mathrm{n}}\mathrm{C}_5}$

$$\begin{gathered} =\underset{\mathrm{n}\to \infty }{\lim }\frac{\mathrm{n}\times \mathrm{n}\left(\mathrm{n}-1\right)\left(\mathrm{n}-2\right)\left(\mathrm{n}-3\right)}{4!}\times \frac{\mathrm{n}(\mathrm{n}-1)(\mathrm{n}-2)}{3!}\times \frac{\mathrm{n}\left(\mathrm{n}-1\right)}{2!}\times \frac{{\left(5!\right)}^2{\left(\left(2\mathrm{n}-5\right)!\right)}^2}{{\left(2\mathrm{n}!\right)}^2} \\ =\underset{\mathrm{n}\to \infty }{\lim }\frac{{\mathrm{n}}^4{(\mathrm{n}-1)}^3{(\mathrm{n}-2)}^2(\mathrm{n}-3){\left(\left(2\mathrm{n}-5\right)!\right)}^2\times 5\times 5\times 4\times 3!}{3!2!{\left(2\mathrm{n}!\right)}^2} \\ =\underset{\mathrm{n}\to \infty }{\lim }\frac{50.{\mathrm{n}}^4{(\mathrm{n}-1)}^3{(\mathrm{n}-2)}^2(\mathrm{n}-3)}{\left(2\mathrm{n}\right(2\mathrm{n}-1\left)\right(2\mathrm{n}-2\left)\right(2\mathrm{n}-3){(2\mathrm{n}-4))}^2}=\frac{50}{1024}=\frac{25}{512} \end{gathered}$$