The arrival rate of customers arriving at a bank counter follows a Poisson distribution with a mean rate of 4 per 10 minutes interval. Find the probability that i. Exactly 0 customer will arrive in 10 minutes interval ii. Exactly 2 customers will arrive in 10 minutes interval iii. At most 2 customers will arrive in 10 minutes interval iv. At least 3 customers will arrive in 10 minutes interval
Let "N" be the number of customers arriving in 10 minutes interval.
"N" has "Poisson(4)" distribution, hence "P(N=n)=\\frac{4^ne^{-4}}{n!}".
i. "P(N=0)=\\frac{4^0e^{-4}}{0!}=e^{-4}".
ii. "P(N=2)=\\frac{4^2e^{-4}}{2!}=8e^{-4}".
iii.
"P(N\\leq2)=\\\\\nP(N=0)+P(N=1)+P(N=2)=\\\\\ne^{-4}+\\frac{4^1e^{-4}}{1!}+8e^{-4}=13e^{-4}."
iv.
"P(N\\geq3)=P(N>2)=1-P(N\\leq2)=1-13e^{-4}."
Answer: i. "e^{-4}\\approx0.0183", ii. "8e^{-4}\\approx 0.1465", iii. "10e^{-4}\\approx 0.1831", iv. "1-13e^{-4}\\approx 0.7619".
Comments
Dear Keshav Goyal, thank you for correcting us.
It would be total of 13e(-4) in the third part because 1e(-4) + 4e(-4) + 8e(-4) = 13e(-4) which is equal to 0.2379.
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