Customers arrive at a shop according to the Poisson distribution with a mean of 10 customers/hour. The manager notes that no customer arrives for the first 3 minutes after the shop opens. The probability that a customer arrives within the next 3 minutes is

0.86

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0.61

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0.39

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0.50

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Solution

The correct option is C 0.39
$(i) Probability of zero in 3 min (λ T = 0.5)$

$P_{1} \to P (0, 3) = \frac{(e x p)^{- 0.5} . (0.5)^{0}}{0!} = 0.606$

$Probability of atleast one customer in next 3 minutes = 1 - 0.606 = 0.39$

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