The probability of a customer arrival at a grocery service counter in any 1 second is equal to 0,1. Assume that customers arrive in a random stream and hence that the arrival any 1 second is independent of any other. Calculate the probability that the first arrival will occur during the third 1-second interval.
"p=0.1 \\\\\n\nq=1-q=0.9"
Geometric probability:
"p(y) = q^{y-1}p \\\\\n\np(3) = 0.9^{3-1} \\times 0.1 = 0.081 = 8.1 \\;\\%"
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