If a point $$A(x, y)$$ lies in the region bounded by the y-axis, straight lines $$2y + x = 6$$ and $$5x - 6y = 30$$, then the probability that $$y < 1$$ is

The given region is bounded by

$$x=0$$

$$x+2y=6$$

and

$$5x-6y=30$$

The vertices of the triangle are obtained as follows:

From

$$x=0,\quad x+2y=6$$

we get

$$y=3$$

So, one vertex is

$$(0,3)$$

From

$$x=0,\quad 5x-6y=30$$

we get

$$y=-5$$

So, another vertex is

$$(0,-5)$$

Now, solving

$$x+2y=6$$

and

$$5x-6y=30$$

we get

$$x=6,\qquad y=0$$

Hence, the third vertex is

$$(6,0)$$

Therefore, total area of the triangle is

$$\frac12\times8\times6=24$$

Now, the line

$$y=1$$

intersects

$$x+2y=6$$

at

$$x=4$$

So, the region where

$$y\ge1$$

is a triangle with vertices

$$(0,3),\ (0,1),\ (4,1)$$

Its area is

$$\frac12\times2\times4=4$$

Hence, area where

$$y<1$$

is

$$24-4=20$$

Therefore, required probability is

$$\frac{20}{24}=\frac56$$

Hence, the correct answer is

$$\boxed{\text{Option B }\frac56}$$