Picky Baggy

Let the number of red balls be $r$ and the number of blue balls be $b$ , with the total being $r+b=n$ balls. Then the probability of choosing a red then a blue would be $\frac{r}{n}\frac{b}{n-1}$ ; likewise, the probability of choosing blue then red would be $\frac{b}{n}\frac{r}{n-1}$ . Adding these together we get the total probability of choosing a different color as

$\frac{2rb}{n(n-1)}$

Setting this equal to $\frac{1}{2}$ and rearranging, we get

$rb=\frac{n(n-1)}{4}$

Since $r$ and $b$ are integers, we must have that $\frac{n(n-1)}{4}$ is also an integer. The smallest $n$ for which this holds is $\boxed{n=4}$ , which satisfies the problem by setting $(r,b)=(1,3)$

Let $R$ be the number of red balls and $B$ the number of blue balls. The probability that drawing two results in 1 red, 1 blue is $\mathbb P = \frac{\binom R 1 \binom B 1}{\binom {R+B} 2} = \frac{RB}{\tfrac12(R+B)(R+B-1)}.$ This should be equal to $\tfrac12$ . Thus $\tfrac12(R+B)(R+B-1) = 2RB$ $R^2+B^2+2RB-R-B = 4RB$ $R^2 + B^2 - 2RB = R + B$ $(R - B)^2 = R+B.$ Thus the difference of $R$ and $B$ is the square of their sum . We try the smallest possible combinations: $1^2 = 1\ \ \ \ \therefore\ \ \ \ R - B = 1, R + B = 1, R = 1, B = 0;$ this fails because there must be at least two balls in the bag. $2^2 = 4\ \ \ \ \therefore\ \ \ \ R - B = 2, R + B = 4, R = 3, B = 1$ showing that we can make this work with $\boxed{\text{four balls}}$ : three red and one blue, or vice versa. The next possible solution would be $3^2 = 9\ \ \ \ \therefore\ \ \ \ R - B = 3, R + B = 9, R = 6, B = 3$ and so on.

Check: $\mathbb P = \frac{\binom 3 1 \binom 1 1}{\binom 4 2} = \frac{3\cdot 1}{6} = \frac12$ as desired.

Are you kidding me? This should be LEVEL 1.

Minimum possible is 4 (as per the options given) as with 4 balls you can visualize an outcome where you take a red one out and then a blue one out or vice versa. You need not perform laborious calculations.