There are 6 cards, 3 blue and 3 red. What is the probability of putting them in a row with alternating colors?

Question

I solved this question this way.

First, there are two ways that the cards will not be alternating:

B - R - B - R - B - R

R - B - R - B - R - B

Second, there are 6! (720) possible orders in which the cards can be dealt.

So, the answer is 2/720. Is this correct?

Answer 1

Without loss of generality, the first card is blue. Then choose a red one with probability $\frac{3}{5}$, a blue one with probability $\frac{2}{4}=\frac{1}{2}$, a red one with probability $\frac{2}{3}$, and the last blue one with probability $\frac{1}{2}$, followed by the last red one. The result is $\frac{2}{5}\cdot(\frac{1}{2})^2=\frac{1}{10}$.

Answer 2

There are actually $6!/(3!3!)=720/36=20$ ways, so the probability is $2/20=1/10$.