Answer:
0.5372
Step-by-step explanation:
Given that the number of births that occur in a hospital can be assumed to have a Poisson distribution with parameter = the average birth rate of 1.8 births per hour.
Let X be the no of births in the hospital per hour
X is Poisson
with mean = 1.8
the probability of observing at least two births in a given hour at the hospital
= 
the probability of observing at least two births in a given hour at the hospital = 0.5372
Solution:
The difference of cubes identity is
if a and b are any two real numbers, then difference of their cubes , when taken individually:
→a³ - b³= (a-b)(a² + a b + b²)→→→Option (D) is true option.
I will show you , how this identity is valid.
Taking R H S
(a-b)(a² +b²+ab)
= a (a² +b²+ab)-b(a² +b²+ab)
= a³ + a b² +a²b -b a² -b³ -ab²
Cancelling like terms , we get
= a³ - b³
= L H S