Question

Airline passengers arrive randomly and independently at the passenger-screening facility at a major international airport. The mean arrival rate is 6 passengers per minute.

A. Compute the probability of no arrivals in a one-minute period (to 6 decimals). B. Compute the probability that three or fewer passengers arrive in a one-minute period (to 4 decimals). C. Compute the probability of no arrivals in a 15-second period (to 4 decimals). Compute the probability of at least one arrival in a 15-second period (to 4 decimals).

Answer 1

here we have,

mean arrival rate, = 6

we know that,

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a) the probability of no arrivals in a one-minute period

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b) the probability that three or fewer passengers arrive in a one-minute period

P(X<=3) = P(X=0) + P(X=1) + P(X=2) + P(X=3)

so,

therefore, P(X<=3) = 0.00248 + 0.01487 + 0.04462 + 0.08924 = 0.1512 (Rounded to 4 decimal places)

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c) here first we have to find arrival rate for 15 seconds

so, arrival rate = (6/60) * 15 = 1.5

the probability of no arrivals in a 15-second period

therefore P(X=0) = 0.2231 (Rounded to 4 decimal places)

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d) the probability of at least one arrival in a 15-second period