Masses' probability

We can have:

$6$ only as ( $1 + 2 + 3$ )

$7$ only as ( $1 + 2 + 4$ )

$8$ only as ( $1 + 2 + 5$ ) and ( $1 + 3 +4$ )

$9$ only as ( $1 + 2 + 6$ ) ( $1 + 3 + 5$ ) ( $2 + 3 + 4$ )

$P_{6} = \frac {1} {8 \cdot 7 \cdot 6} \cdot 3!$

$P_{7} = P_{6}$

$P_{8} = 2P_{6}$

$P_{9} = 3P_{6}$

$P _{tot} = P _{6} + P _{7} + P _{8} + P _{9} = \frac 18$

There are ${ C }_{ 8 }^{ 3 }=56$ ways to choose randomly $3$ weights from $8$ weights.

There are also $7$ outcomes showing that sum of $3$ chosen weights is not greater than $9kg$ as below:

$\left( 1,2,6 \right) ,\left( 1,3,5 \right) ,\left( 2,3,4 \right) ,\left( 1,2,3 \right) ,\left( 1,2,4 \right) ,\left( 1,2,5 \right) ,\left( 1,3,4 \right)$

Hence, the probability that sum of $3$ chosen weights is not greater than $9kg$ is: $P=\frac { 7 }{ 56 } =\frac { 1 }{ 8 } =0.125$ .