Which of the following cannot be a valid assignment of probabilities for outcomes of sample space $S = \{\omega_{1}, \omega_{2}, \omega_{3}, \omega_{4}, \omega_{5}, \omega_{6}, \omega_{7}\}$?

Outcome	Probability
$\omega_{1}$	$\frac{1}{14}$
$\omega_{2}$	$\frac{2}{14}$
$\omega_{3}$	$\frac{3}{14}$
$\omega_{4}$	$\frac{4}{14}$
$\omega_{5}$	$\frac{5}{14}$
$\omega_{6}$	$\frac{6}{14}$
$\omega_{7}$	$\frac{15}{14}$

(D) For any probability assignment to be valid,it must satisfy two conditions:
$1$. Each probability $P(\omega_{i})$ must be such that $0 \leq P(\omega_{i}) \leq 1$.
$2$. The sum of all probabilities must be equal to $1$,i.e.,$\sum P(\omega_{i}) = 1$.
In the given table,we observe the probability assigned to $\omega_{7}$ is $P(\omega_{7}) = \frac{15}{14}$.
Since $\frac{15}{14} > 1$,this violates the fundamental axiom of probability that states $P(\omega_{i}) \leq 1$ for all $i$.
Therefore,this assignment is not valid.