Answer:
c) 
Step-by-step explanation:
Rate of collision,
1.2 collisions every 4 months
or,
= 0.3 collisions per month
So, the Poisson distribution for the random variable no. of collisions per month (X) is given by,
P(X =x) = 
for x ∈ N ∪ {0}
= 0 otherwise --------------------------------------(1)
here,
collision / month
No collision over a 4 month period means no collision per month or X =0
Putting X = 0 in (1) we get,
P(X = 0) = 
------------------------------------(2)
Now, since we are calculating this for 4 months,
so, P(No collision in 4 month period)
=
-----------------------------------------------------------(3)
2 collision in 2 month period means 1 collision per month or X =1
Putting X =1 in (1) we get,
P(X =1) = 
------------------------------------(4)
Now, since we are calculating this for 2 months, so ,
P(2 collisions in 2 month period)
=
-----------------------------------------(5)
1 collision in 6 months period means
collision per month
Now, P(1 collision in 6 months period)
= P( X = 1/6] (which is to be estimated)
=
= \frac {0.7408182207 \times 5 + 0.2222454662 \times 1}{6}[/tex]
-------------------------------------------(6)
So,
P(1 collision in 6 month period)
= 
------------------------------------------------(7)
So,
P(No collision in 6 months period)
= 
---------------------------------(8)
so,
P(1 or fewer collision in 6 months period)
= (8) + (7 ) = 0.0785267444 +0.1652988882
---------------------------------------------(9)