Let X be the number of lightning strikes in a year at the top of particular mountain.
X follows Poisson distribution with mean μ = 3.8
We have to find here the probability that in randomly selected year the number of lightning strikes is 0
The Poisson probability is given by,
P(X=k) = [tex] \frac{e^{-mean} mean^{x}}{x!} [/tex]
Here we have X=0, mean =3.8
Hence probability that X=0 is given by
P(X=0) = [tex] \frac{e^{-3.8} 3.8^{0}}{0!} [/tex]
P(X=0) = [tex] \frac{0.02237 * 1}{1} [/tex]
P(X=0) = 0.0224
The probability that in a randomly selected year, the number of lightning strikes is 0 is 0.0224
