imc

There are two ways to arrange the red beads, where $R$ represents a red bead and $-$ represents a blank space. $1.R-R-R-$ $2.R--R-R$ In the first, there are three ways to place a bead in the first free space, two for the second free space, and one for the third, so there are $6$ arrangements. In the second, a white bead must be placed in the third free space, so there are two possibilities for the third space, two for the second, and one for the first. That makes $4$ arrangements. There are $6+4=10$ arrangements in total. The two cases above can be reversed, so we double $10$ to get $20$ arrangements. Also, in each case, there are three ways to place the first red bead, two for the second, and one for the third, so we multiply by $6$ to get $20\cdot 6=120$ arrangements. There are $6!=720$ total arrangements so the answer is $120/720=\boxed{1/6}$.