The probability that a ticket holder will miss a flight is .005. Of 180 passengers take the flight, what is the approximate Poisson probability that at least 2 will miss the flight

The Poisson probability that at least 2 will miss the flight P(x ≥ 2) is 0.22753

Step-by-step explanation:

Given that;

probability that a ticket holder will miss a flight = 0.005

(number of passengers) sample size n = 180

so avg. number of passengers missing the flight will be λ = 0.005 × 180 = 0.9

so X: number of passengers will miss the flight

where x follows Poisson distribution with λ as 0.9

Probability mass function of Poisson  is expressed as;

P(x) = (e^-λ)λˣ / x!

so we substitute

P(x) = (e^-0.9)0.9ˣ / x!

Now Probability that at least 2 will miss the flight;

P(x ≥ 2) = 1 - P(x < 2) = 1 - (P(x=0) + P(x=1) )

so

for P(x=0); P(0) = (e^-0.9)0.9⁰ / 0! = 0.40656

for P(x=1); P(1) = (e^-0.9)0.9¹ / 1! =  0.36591

hence;

(P(x=0) + P(x=1) )  = 0.40656 + 0.36591 = 0.77247

P(x ≥ 2) = 1 - P(x < 2) = 1 - (P(x=0) + P(x=1) )

P(x ≥ 2) =  1 - 0.77247 = 0.22753

Therefore, The Poisson probability that at least 2 will miss the flight P(x ≥ 2) is 0.22753

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