91Ó°ÊÓ

A drugstore has 3 public telephone booths.\({p_i}\)denotes the probability that exactly i^th telephone booth will be occupied on any Monday evening at 8.00 p.m. where\(i = 0,1,2,3\).

The probabilities are\({p_0} = 0.1,{p_1} = 0.2,{p_2} = 0.4\;and\;{p_3} = 0.3\)

X and Y are the numbers of booths occupied at 8.00 p.m. On two Monday evenings independently. So, X and Y are independent.

The function of joint probability distribution X, as well as Y, is

\(\begin{array}{c}\Pr \left( {{x_i},{y_j}} \right) = \Pr \left( {X = {x_i},Y = {y_j}} \right)\\ = \Pr \left( x \right)\Pr \left( y \right)\\ = {P_x}{P_y}\left( {say} \right)\end{array}\)

Thus, the function of joint probability X, as well as Y, is

\(\Pr \left( {x,y} \right) = \left\{ \begin{array}{l}{p_x}{p_y}\;for\;x = y = 0,1,2,3\\0\;otherwise\end{array} \right.\)

Referring to the part a. from the calculated joint distribution, we get the probabilities,

Now from the probabilities we calculated above, we can determine the required probability,

\(\begin{array}{c}\Pr \left( {X = Y} \right) = \left( \begin{array}{l}\Pr \left( {X = 0\left| {Y = 0} \right.} \right) + \Pr \left( {X = 1\left| {Y = 1} \right.} \right)\\ + \Pr \left( {X = 2\left| {Y = 2} \right.} \right) + \Pr \left( {X = 3\left| {Y = 3} \right.} \right)\end{array} \right]\\ = 0.01 + 0.04 + 0.16 + 0.09\\ = 0.30\end{array}\)

Thus, \(\Pr \left( {X = Y} \right) = 0.30\).

Referring to the above table from part b, we calculate the required probability.

So,

\(\begin{array}{c}\Pr \left( {X > Y} \right) = \left( \begin{array}{l}\Pr \left( {X = 1\left| {Y = 0} \right.} \right) + \Pr \left( {X = 2\left| {Y = 0} \right.} \right) + \\\Pr \left( {X = 2\left| {Y = 1} \right.} \right) + \Pr \left( {X = 3\left| {Y = 0} \right.} \right) + \\\Pr \left( {X = 3\left| {Y = 1} \right.} \right) + \Pr \left( {X = 3\left| {Y = 2} \right.} \right)\end{array} \right)\\ = 0.02 + 0.04 + 0.08 + 0.03 + 0.06 + 0.12\\ = 0.35\end{array}\)

Thus,\(\Pr \left( {X > Y} \right) = 0.35\).