Poisson distribution problem

[rayne1]

Problem:
McBurger’s drive-thru has only one service window and serves an average of 2 customers every 5 minutes. 70% of customers order drinks from the drive-thru.

The manager monitors the employee at the drive-thru for the next 3 hours. He will give the employee a raise if exactly 20 customers are served per hour in at least 1 of these hours. What is the probability that the employee will receive a raise?
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I know the probability that there are exactly 20 customers in the next hour is 0.0624 (rounded), but don't know how to deal with the "at least 1 of these (3) hours" part of the question.

 

[I like Serena]

Hi rayne!

Hint: what is the chance that none of the 3 hours satisfies the criterion?

 

[rayne1]

Hi rayne!

Hint: what is the chance that none of the 3 hours satisfies the criterion?

Not sure :(

 

[I like Serena]

Not sure :(

A property of the Poisson distribution is that the events in each time interval are independent from each other.

And one of the properties of probabilities is that if $A,B,C$ pairwise independent, then:
$$P(A\wedge B\wedge C) = P(A)P(B)P(C)$$

 

[rayne1]

A property of the Poisson distribution is that the events in each time interval are independent from each other.

And one of the properties of probabilities is that if $A,B,C$ pairwise independent, then:
$$P(A\wedge B\wedge C) = P(A)P(B)P(C)$$

Okay, but I'm still unclear on how to solve it.