Suppose you have $6$ fair weighted dice with all faces having $1$ to $6$. since the outcome is mutually independent and are disjointed sets, the probability of getting at least one six is $P(a)=6\cdot(1/6)$ by law of addition but if now there are $7$ dices so $P(a)$ is $7\cdot(1/6)$,how?
Now a follow up question -what is the probability of $6$ dices having Same sides except $6$ which is $P(b)=5(\frac16)^6$
But we know $P(a)$, probabability of getting at least $1$ six in $6$ dice is $1$, so this means $P(b)=0$ since one side must have six but earlier we exclude it.
The probability of having at least one 6 is $1-(\tfrac{5}{6})^6$.
The events are not disjoint: you can have 6 in dice #1 and in dice #2 at the same time.