Answer:
[tex]P(X = 0) = 0.9891[/tex]
Step-by-step explanation:
Given
[tex]\lambda = 0.011[/tex]
Required [This completes the question]
The probability of exactly 0 deaths
This probability follows a Poisson distribution, and it is given by:
[tex]P(X = x) = \frac{e^{-\lambda}\lambda^{x}}{x!}[/tex]
For 0 deaths;
[tex]x = 0[/tex]
So, the expression becomes
[tex]P(X = 0) = \frac{e^{-\lambda}\lambda^{0}}{0!}[/tex]
[tex]P(X = 0) = \frac{e^{-\lambda}\lambda^{0}}{1}[/tex]
[tex]P(X = 0) = \frac{e^{-\lambda}*1}{1}[/tex]
[tex]P(X = 0) = e^{-\lambda}[/tex]
Substitute 0.011 for [tex]\lambda[/tex]
[tex]P(X = 0) = e^{-0.011}[/tex]
[tex]P(X = 0) = 0.9891[/tex]
The probability of having exactly death is 0.9891
