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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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up vote 3 down vote accepted

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 $P(A_5 \cap A_6)$?

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P(A5 U A6) = P(A5)+P(A6) - P(A5 intersect A6), so I just solve for the intersection? But I'm not clear whether I'm being asked the probability of at least one 5 and 6. – lord12 Sep 14 '11 at 3:02
It's always a good idea to be clear about exactly what question you are being asked. If necessary, ask the instructor for a clarification. – Robert Israel Sep 14 '11 at 17:59

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 $6$?

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I wanted the probability of outcomes 5 and 6 both occur at least once. n and k are the same. – lord12 Sep 14 '11 at 2:58

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