Question

In each of the months June, July, and August, the number of accidents occurring in that month is modeled by a Poisson random variable with mean 1. In each of the other 9 months of the year, the number of accidents occurring is modeled by a Poisson random variable with mean 0.5. Assume that these 12 random variables are mutually independent. Calculate the probability that exactly two accidents occur in July through Novembe

Answer 1

Answer:

$P(X=2)=0.1849$

Step-by-step explanation:

From the question we are told that

Sample size n=3

Sample mean 1  $\=x_1=1$

Sample mean 2 $\=x_2=0.5$

Generally the Probability that 2 accidents occurs from July to August is mathematically given by

 $mean\ accidents=2*\=x_1$

 $mean\ accidents=2*1$

 $mean\ accidents=2$

Generally the probability that 2 deaths occurs from September through November is mathematically given by

 $Mean\ accidents=3*\=x_2$

 $Mean\ accidents=3*0.5$

 $Mean\ accidents=1.5$

Therefore

Total Mean accidents $T_a=2+1.5=>3.5$

Generally the probability of  two accident per month is mathematically given by

$P(X=2)=\frac{e^-^3^.^5*3.5^2}{2!}$

$P(X=2)=0.1849$

Therefore the probability of having  2 accidents from July through to November is

$P(X=2)=0.1849$