Question

On the average a submarine on patrol sights $6$ enemy ships per hour. Assuming the number of ships sighted in a given length of time is a poisson variate, the probability of sighting $4$ ships in the next two hours is

• A. $\frac{e^{-12}{12}^4}{4!}$
• B. $\frac{e^{-4}{12}^{12}}{3!}$
• C. $\frac{e^{-6}{12}^4}{4!}$
• D. $\frac{e^{-3}{12}^2}{4!}$

Answer 1

The correct option is A $\frac{e^{-12}{12}^4}{4!}$

By using Poisson distribution,
$P(x;\mu )=\frac{e^{-\mu }{\mu }^x}{x!}$
$\mu$: The mean number of successes that occur in a specified region(parameter)
x: The actual number of successes that occur in a specified region.
P(x; $\mu$): The Poisson probability that exactly x successes occur in a Poisson experiment, when the mean number of successes is $\mu$
the average a submarine on patrol sights 6 enemy ships per hour, so for 2 hours
Here, $\mu =6\times 2=12$
x = 4 ( ships)
$P(4;12)=\frac{e^{-12}{12}^4}{4!}$