Answer:
[tex]a. \ P(X=0)=5.574\times10^{-7}[/tex]
[tex]b. \ \ P(X\leq 1)=8.584\times10^{-6}[/tex]
[tex]c.\ \ \ P(X\geq 4)=0.9997[/tex]
Step-by-step explanation:
a. #We notice this is a Poisson probability function expressed as:
[tex]f(x)=\frac{\mu^xe^{-\mu}}{x!}[/tex]
x-number of occurrences in a given interval.
[tex]\mu[/tex]-mean occurrences of the event
-The mean is calculated as:
[tex]\mu=\frac{14.4}{4}=3.6[/tex]
#the probability of no accidents in a 15-minute period is :
[tex]P(x)=\frac{\mu^xe^{-\mu}}{x!}\\\\P(X=0)=\frac{14.4^0e^{-14.4}}{0!}\\\\=5.574\times10^{-7}[/tex]
Hence, the probability of no accident in a 15-min period is [tex]=5.574\times10^{-7}[/tex]
b. The the probability of at least one accident in a 15-minute period. is calculated as:
[tex]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}[/tex]
Hence, the probability of at least one accident in a 15-minute period is [tex]8.584\times10^{-6}[/tex]
c. The probability of four or more accidents in a 15-minute period is calculated as:
[tex]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[/tex]
Hence,the probability of four or more accidents in a 15-minute period. is 0.9997
