binomial probability question

If you have k trials that result in 5,6,7 with probabilities $P(5)$, $P(6)$, and $P(7)$ (respectively), with

$P(5) + P(6) + P(7) = 1$,

what is the probability that 5 and 6 occur at least once.

Would it be $1-P(\text{5 and 6 never occur}) = 1- (P(5)+P(6))^0(P(7))^n = 1 - P(7)^n$?

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I think you mean the probability that there is at least one 5 and at least one 6 in $k$ independent trials. Let $A_5$ be the event that there is at least one 5, $A_6$ the event that there is at least one 6. The probability that there is at least one 5 or at least one 6 is $P(A_5 \cup A_6) = 1 - p_7^k$, while $P(A_5) = 1 - (p_6 + p_7)^k$ and $P(A_6) = 1 - (p_5 + p_7)^k$. Now, do you know an equation that would express $P(A_5 \cup A_6)$ in terms of $P(A_5)$, $P(A_6)$ and $P(A_5 \cap A_6)$?
I assume $k$ and$n$ are the same. $1-p_7^k$ is the chance of at least one trial not being $7$, so at least one $5$ or one $6$. Is that what you wanted, or the probability of at least one $5$ and at least one $6$?