Question 353740
GIVEN: an incident of shoplifting caught by security 0.8 times every five hours
This implies that on average, 0.8/5=0.16 incidents caught 
Over 11 hours, the average incidents caught is 11*0.16=1.76
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This problem is Poisson because there are shoplifting incidents or there are not and each incident of being caught is very small, while the potential for shoplifiting incidents can be large  (n-->large P-->0) with an average of 1.76 over a period of 11 hours.
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(b) What is the probability that from 10 A.M. to 9 P.M. there will be at least one shoplifting incident caught by security? (Use 4 decimal places.)
NOTE:  your problems began with b)
let X=number of incidents caught
P(X>=1)=1-P(X=0)={{{1-e^(-1.76)=1-0.1720=0.8280}}}
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(c) What is the probability that from 10 A.M. to 9 P.M. there will be at least three shoplifting incidents caught by security? (Use 4 decimal places.)
P(X>=3)=1-P(X<=2)=1-p(X=0)-P(X=1)={{{1-0.1720-1.76*e^(-1.76)=1-0.1720-0.30280=0.5252}}}
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(d) What is the probability that from 10 A.M. to 9 P.M. there will be no shoplifting incidents caught by security? (Use 4 decimal places
{{{P(X=0)=e^(-1.76)=0.1720}}}