How likely?

Clearly a boy must be first in the queue, for if a girl was first then they would have an equal number of boys and girls, (namely $0$ ), ahead of them.

Now with one of the boys at the head of the queue and one of the girls in the second position, there would have to be a boy in the third position, since otherwise a girl in the third position would have an equal number of boys and girls ahead of them, (namely $1$ each). Then with a boy in the third position, the second girl can go in either the fourth or fifth position and have the conditions of the problem satisfied.

With two boys at the head of the queue and a girl in the third position, the second girl can go in either the fourth or fifth position and have more boys ahead of them than girls.

Finally, having the girls in the fourth and fifth positions would satisfy the conditions as well. This gives us a total of $5$ generic sequences, namely

$BGBGB, BGBBG, BBGGB, BBGBG, BBBGG$ ,

that satisfy the conditions of the problem. As there are a total of $\dfrac{5!}{3!*2!} = 10$ possible generic sequences without restrictions, the desired probability is $\dfrac{5}{10} = \boxed{0.50}$ .

Note that by "generic" sequence I mean a sequence without regard for which boys and which girls fill each specific spot. Since every generic sequence has $5!$ ways in which it can be realized, and we are looking for a probability rather the number of possible sequences, there is no need to worry about the specific placements of boys and girls.

For a very fast and incomplete answer, notice that the total number of ways to choose the girls' positions is $\binom{5}{2} = 10$ So our answer will be a multiple of $\frac{1}{10}.$ The only answer out of those given that is a multiple of $0.1$ is $\boxed{0.50}$