[Math] Probability of boys ahead girls

Question

There are three boys and two girls in a Queue. What is the Probability that number of boys ahead of every girl is at least one more than number of girls ahead of her.

My Try: Let $B1$,$B2$,$B3$ are Boys and $G1$,$G2$ are Girls

The Possible Outcomes are:

$1.$ $G1$ $G2$ $B1$ $B2$ $B3$

$2.$ $G1$ $B1$ $G2$ $B2$ $B3$

$3.$ $G1$ $B1$ $B2$ $G3$ $B3$

$4.$ $B1$ $G1$ $G2$ $B2$ $B3$

$5.$ $B1$ $G1$ $B2$ $G2$ $B3$

So The Required Probability is $$\frac{5\times2!\times3!}{5!}=\frac{1}{2}$$ Please let me know if there is better way to do this..

Answer 1

Not much quicker, but you could look at patterns that fail

. . . . G with prob 2/5
B B G G B with prob 1/10

and $1-\left(\frac25 +\frac1{10}\right) = \frac12$