Question

On an 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 at least two ships in the next $20$ minutes is

• A. $1-e^{-2}$
• B. $1-2e^{-2}$
• C. $1-3e^{-2}$
• D. $1-4e^{-2}$

Answer 1

The correct option is C $1-3e^{-2}$

The probability of sighting at least two ships
=1-(Probability of sighting atmost 1 ships)
$=1-e^{-\lambda }-\frac{{\lambda }^1{e}^{-\lambda }}{1!}$
Here $\lambda =2$
Hence
$P=1-e^{-\lambda }-\frac{{\lambda }^1{e}^{-\lambda }}{1!}$
$=1-e^{-2}-2e^{-2}$
$=1-3e^{-2}$.