Answer:
21.77% probability that the antenna will be struck exactly once during this time period.
Explanation:
In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:
![P(X = x) = (e^(-\mu)*\mu^(x))/((x)!)](https://img.qammunity.org/2021/formulas/mathematics/college/frjienvs346ki5axyreyxszxd4zhu8xxhm.png)
In which
x is the number of sucesses
e = 2.71828 is the Euler number
is the mean in the given interval.
In this question:
![\mu = 2.40](https://img.qammunity.org/2021/formulas/mathematics/college/30qk7ruh7bb2yliug7q7f5n8bz8i0n7txt.png)
Find the probability that the antenna will be struck exactly once during this time period.
This is P(X = 1).
![P(X = x) = (e^(-\mu)*\mu^(x))/((x)!)](https://img.qammunity.org/2021/formulas/mathematics/college/frjienvs346ki5axyreyxszxd4zhu8xxhm.png)
![P(X = 1) = (e^(-2.40)*2.40^(1))/((1)!) = 0.2177](https://img.qammunity.org/2021/formulas/mathematics/college/wdm2gcsftmwjgu3a7ihc9hmufhrqvjufdr.png)
21.77% probability that the antenna will be struck exactly once during this time period.