An urn contains r red balls and b black balls

	COLUMN – I		COLUMN – II
A)	If the probability of getting two Red balls in first two draws (Without replacement) is 1/2, Then value of r can be	p)	10
B)	If the probability of getting two red Balls in first two draws (Without replacement) is 1/2 and b is an even number, then r can be	q)	3
C)	If the probability of getting exactly two red balls in four draws (with replacement) from the urn is 3/8 and b = 10, then r can be	r)	8
D)	If the probability of getting exactly n red balls in 2n draws (with replacement) is equal to probability of getting exactly n black balls in 2n draws (with replacement), then the ratio r/b can be	s)	2

$\begin{array}{r}\text{ A) }\left(\frac{\mathrm{r}}{\mathrm{r}+\mathrm{b}}\right)\cdot \left(\frac{\mathrm{r}-1}{\mathrm{r}+\mathrm{b}-1}\right)=\frac{1}{2}\\ \Rightarrow {\mathrm{b}}^2+(2\mathrm{r}-1)\mathrm{b}+\left(\mathrm{r}-{\mathrm{r}}^2\right)=0\end{array}$
For r = 10, 8, 2; $\mathrm{\Delta }>0$but not a perfect square For r = 3 is a possible value
B)$\left(\frac{\mathrm{r}}{\mathrm{r}+\mathrm{b}}\right)\left(\frac{\mathrm{r}-1}{\mathrm{r}+\mathrm{b}-1}\right)=\frac{1}{6}$
$\Rightarrow {\mathrm{b}}^2+(2\mathrm{r}-1)\mathrm{b}+\left(5\mathrm{r}-5{\mathrm{r}}^2\right)=0$
For r = 10, 3, 8 ; $\mathrm{\Delta }>0$but not a perfect square For r = 2 solving we get b = 2 (even). 
$\therefore$r = 2 is a possible value
C) Let X be the no. of red balls obtained, out of 4 balls drawn with replacement
$\mathrm{x}~\mathrm{B}(\mathrm{n}=4,\mathrm{p})$Where$\mathrm{p}=\frac{\mathrm{r}}{\mathrm{r}+10},\mathrm{q}=\frac{10}{\mathrm{r}+10}$
$\mathrm{P}(\mathrm{X}=2)=\frac{3}{8}{\Rightarrow }^4{\mathrm{C}}_2{\left(\frac{10}{\mathrm{r}+10}\right)}^2{\left(\frac{\mathrm{r}}{\mathrm{r}+10}\right)}^2=\frac{3}{8}$
Solving we get r = 10
D) Let X be the no. of red balls and y be the no. of black balls obtained in 2n draws with replacement
Then  $\mathrm{X}~\mathrm{B}\left(2\mathrm{n},{\mathrm{P}}_1=\frac{\mathrm{b}}{\mathrm{r}+\mathrm{b}}\right)\mathrm{X}~\mathrm{B}\left(2\mathrm{n},{\mathrm{P}}_1=\frac{\mathrm{b}}{\mathrm{r}+\mathrm{b}}\right)$
And $\mathrm{Y}~\mathrm{B}\left(2\mathrm{n},{\mathrm{P}}_2=\frac{\mathrm{b}}{\mathrm{r}+\mathrm{b}}\right)$
$\begin{array}{l}\mathrm{P}(\mathrm{X}=\mathrm{n}){=}^{2\mathrm{n}}{\mathrm{C}}_{\mathrm{n}}{\left(1-\frac{\mathrm{r}}{\mathrm{r}+\mathrm{b}}\right)}^{\mathrm{n}}{\left(\frac{\mathrm{b}}{\mathrm{r}+\mathrm{b}}\right)}^{\mathrm{n}}\\ \Rightarrow \mathrm{P}(\mathrm{Y}=\mathrm{n}){=}^{2\mathrm{n}}{\mathrm{C}}_{\mathrm{n}}{\left(1-\frac{\mathrm{r}}{\mathrm{r}+\mathrm{b}}\right)}^{\mathrm{n}}{\left(\frac{\mathrm{b}}{\mathrm{r}+\mathrm{b}}\right)}^{\mathrm{n}}\end{array}$
Clearly $\mathrm{P}(\mathrm{X}=\mathrm{n})=\mathrm{P}(\mathrm{Y}=\mathrm{n})\mathrm{\forall }\mathrm{r},\mathrm{b} \mathrm{hence} \frac{\mathrm{r}}{\mathrm{b}} \mathrm{canbe}$
10 or 3 or 8 or 2.