Answer to Question #219516 in Statistics and Probability for mimi

Question

Let the number of rainy days per month in a town has Poisson distribution with a mean of 3. Find the probability of the following events:

(a) There will be one rainy day in one month. (2)

(b) There will be at least two rainy days in one month. (6)

(c) There will be no rain for three consecutive months. (4)

(d) One month of no rain is followed by one month during which it rains on 10 days.

Answer 1

(a) $P(X=1)=e^{-3}\frac{3^1}{1!}=0.1494.$

(b) $P(X\ge2)=1-P(X=0)-P(X=1)=1-e^{-3}\frac{3^0}{0!}-e^{-3}\frac{3^1}{1!}=0.8009.$

(c) This is a binomial distribution with $p=P(X=0)=0.0498,\;n=3.$

$P(X=0)=p^n=0.0498^3=0.0001.$

(d) $P(X=0)=e^{-3}\frac{3^0}{0!}=0.0498.$

$P(X=10)=e^{-3}\frac{3^{10}}{10!}=0.0008.$

$P=0.0498*0.0008=0.00004.$