91Ó°ÊÓ

It is given that X and Y are independent Poisson variables representing the arrival rate of men and women per hour, respectively.

It is also given that

\(\begin{array}{l}X \sim P\left( {120} \right)\\Y \sim P\left( {60} \right)\end{array}\)

Since,

\(X \sim P\left( {120} \right)\,\,and\,\,Y \sim P\left( {60} \right)\)

Let us define a new variable, W=X+Y.

By the property that the sum of the independent Poisson with parameters\(\lambda \,\,and\,\,\mu .\)is a Poisson with parameter\(\lambda \,\, + \,\,\mu .\)

\(\begin{array}{l}W \sim P\left( {120 + 60} \right)\\W \sim P\left( {180} \right)\end{array}\)

We convert the parameter rate into minutes

\(\begin{array}{l}{\rm{For}}\,\,{\rm{W,}}\\{\rm{ = 180}}\,\,{\rm{per}}\,\,{\rm{hour}}\\{\rm{ = 3}}\,\,{\rm{per}}\,\,{\rm{minute}}\end{array}\)

\(\begin{array}{l} = \Pr \left( {W \le 4} \right)\\ = \Pr \left( {W = 0} \right) + \Pr \left( {W = 1} \right) + \Pr \left( {W = 2} \right) + \Pr \left( {W = 3} \right) + \Pr \left( {W = 4} \right)\\ = \frac{{{e^{ - 3}}{3^0}}}{{0!}} + \frac{{{e^{ - 3}}{3^1}}}{{1!}} + \ldots + \frac{{{e^{ - 3}}{3^4}}}{{4!}}\\ = 0.8152.\end{array}\)

Therefore, the final answer is 0.8152