One partition total normalizes three explicit state weights. Exact arithmetic here means exact results for the stated model inputs; measured inputs still carry uncertainty and significant-figure limits.

highlighted = computed this step

State row 1 uses its weight over Z

State 0 has weight 4 over partition total 7, giving probability 4/7.

P0=47=47P_{0}=\frac{4}{7}=\frac{4}{7}
Ensemble probability row 1The same ensemble source supplies every state row.state=0, weight=4, P=4/7

State row 2 uses its weight over Z

State 1 has weight 2 over partition total 7, giving probability 2/7.

P1=27=27P_{1}=\frac{2}{7}=\frac{2}{7}
Ensemble probability row 2The same ensemble source supplies every state row.state=1, weight=2, P=2/7

State row 3 uses its weight over Z

State 2 has weight 1 over partition total 7, giving probability 1/7.

P2=17=17P_{2}=\frac{1}{7}=\frac{1}{7}
Ensemble probability row 3The same ensemble source supplies every state row.state=2, weight=1, P=1/7

The same partition total normalizes each state

The denominator stays seven across all rows. Larger explicit weight gives the larger probability.

statewZP047471272721717\begin{array}{c|c|c|c}state&w&Z&P\\0&4&7&\frac{4}{7}\\1&2&7&\frac{2}{7}\\2&1&7&\frac{1}{7}\\\end{array}
Ensemble probability cross-scanThe source check still carries the full weighted ensemble.weights 4, 2, 1 normalize through one Z=7