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Mathematics
11.04.2023 19:59
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A small commuter plane has 30 seats. The probability that any particular passenger

A small commuter plane has 30 seats. The probability that any particular passenger will not show up for a flight is 0.10, independent of other passengers. The airline sells 32 tickets for the flight. (a) (3pts) Let X be the number of no show. What are possible values of X
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luisr0754
luisr0754
4,9(82 marks)

a. \ P(X=0)=5.574\times10^{-7}

b. \ \ P(X\leq 1)=8.584\times10^{-6}

c.\ \ \ P(X\geq 4)=0.9997

Step-by-step explanation:

a. #We notice this is a Poisson probability function expressed as:

f(x)=\frac{\mu^xe^{-\mu}}{x!}

x-number of occurrences in a given interval.

\mu-mean occurrences of the event

-The mean is calculated as:

\mu=\frac{14.4}{4}=3.6

#the probability of no accidents in a 15-minute period is :

P(x)=\frac{\mu^xe^{-\mu}}{x!}\\\\P(X=0)=\frac{14.4^0e^{-14.4}}{0!}\\\\=5.574\times10^{-7}

Hence, the probability of no accident in a 15-min period is =5.574\times10^{-7}

b. The the probability of at least one accident in a 15-minute period. is calculated as:

P(x)=\frac{\mu^xe^{-\mu}}{x!}\\\\P(X\leq 1)=\frac{14.4^0e^{-14.4}}{0!}+\frac{14.4^1e^{-14.4}}{1!}\\\\=5.574\times10^{-7}+8.026\times 10^{-6}\\\\=8.584\times 10^{-6}

Hence,  the probability of at least one accident in a 15-minute period is 8.584\times10^{-6}

c. The probability of four or more accidents in a 15-minute period is calculated as:

P(X\geq 4)=1-P(X\leq 3)=1-[P(X=0)+P(X1)+P(X=2)+P(X=3)]\\\\=1-[5.574\times10^{-7}+a. \ 8.026\times10^{-6}+\frac{14.4^2e^{-14.4}}{2!}+\frac{14.4^3e^{-14.4}}{3!}]\\\\=1-[8.584\times 10^{-6}+5.779\times10^{-5}+2.774\times10^{-4}]\\\\=0.9997

Hence,the probability of four or more accidents in a 15-minute period. is  0.9997

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