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An own contains r red balls and b black balls

COLUMN-I

COLUMN-II
I)
If the probability of getting two red balls
in the 1^st two draws (without
replacement) is $\frac{1}{2}$ then the value of r can be
P) 10
II)
If the probability of getting two red balls
in the 1^st two draws (without
replacement) is $\frac{1}{2}$ and b is even number,
then r can be
Q) 3
III)
If the probability of getting exactly two
red balls in four draws (with
replacement) from the urn is $\frac{3}{8}$ and
$b = 10$ , then r can be
R) 8
IV)
If the probability getting exactly $n$ red
balls in $2 n$ draws (with replacement) is
equal to probability of getting exactly $n$ black balls in $2 n$ draws (with replacement) then the ratio of $\frac{r}{b}$ can be
S) 2

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$A = Q, B - S, C - P R, D - P Q R$

$A = Q, B - S, C - P Q R, D - P R$

$A = Q, B - S, C - P, D - P Q R S$

$A = Q R, B - S, C - P Q R, D - P R$

answer is A.

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Detailed Solution

$A (\frac{r}{r + b}) (\frac{r - 1}{r + b - 1}) = \frac{1}{2} \Rightarrow b^{2} + (2 r - 1) b + (r - r^{2}) = O$
For $r = 10, 8, 2; Δ > o,$ but not a perfect square
$∴$ r = 3 is possible value
B) $\begin{array}{l} (\frac{r}{r + b}) (\frac{r + 1}{r + b - 1}) = \frac{1}{6} \\ \to b^{2} + (2 r - 1) b + (5 r - 5 r^{2}) = 0 \\ b = 2 \Rightarrow r = 2 \end{array}$
C)

$\begin{array}{l} X ~ B (n = 4, p), p = \frac{r}{r + 10}, 2 = \frac{10}{r + 10} \\ p (x = 2) = \frac{3}{8} \Rightarrow^{4} C_{2} {(\frac{10}{r + 10})}^{2} {(\frac{r}{r + 10})}^{2} = \frac{3}{8} \Rightarrow r = 10 \\ D) X ~ B (2 n, P_{1} = \frac{b}{r + b}) \\ y ~ B (2 n, p_{2} = \frac{b}{r + b}) \\ \frac{r}{b} c a n b e a n y v a l u e p (x = n) = p (y = n) \\ ∴ 10, 8, 3, 2 \end{array}$

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