MHB What is the Probability of Getting a Raise at McBurger's Drive-Thru?

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The discussion centers on calculating the probability of an employee at McBurger's drive-thru receiving a raise based on customer service metrics. The manager will grant a raise if exactly 20 customers are served in at least one of three monitored hours. The probability of serving exactly 20 customers in one hour is established as 0.0624. To find the probability of achieving this in at least one of the three hours, the participants discuss using the complement rule, focusing on the independence of events in the Poisson distribution. The conversation highlights the need to calculate the probability of not meeting the criterion in all three hours to derive the final answer.
rayne1
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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.
 
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Hi rayne!

Hint: what is the chance that none of the 3 hours satisfies the criterion?
 
I like Serena said:
Hi rayne!

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

Not sure :(
 
rayne said:
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:
$$P(A\wedge B\wedge C) = P(A)P(B)P(C)$$
 
I like Serena said:
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.
 

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