Question

Question:A certain drugstore has three public telephone booths. Fori=0, 1, 2, 3, let\({{\bf{p}}_{\bf{i}}}\)denote the probability that exactlyitelephone booths will be occupied on any Monday evening at 8:00 p.m.; and suppose that\({{\bf{p}}_{\bf{0}}}\)=0.1,\({{\bf{p}}_{\bf{1}}}\)=0.2,\({{\bf{p}}_{\bf{2}}}\)=0.4, and\({{\bf{p}}_{\bf{3}}}\)=0.3. LetXandYdenote the number of booths that will be occupied at 8:00 p.m. on two independent Monday evenings. Determine:

(a) the joint p.f. ofXandY;

(b) Pr(X=Y);

(c) Pr(X > Y ).

Short answer

1. The joint p.f of X and Yis\(\Pr \left( {x,y} \right) = \left\{ \begin{array}{l}{p_x}{p_y}\;for\;x = y = 0,1,2,3\\0\;otherwise\end{array} \right.\)
2. \(\Pr \left( {X = Y} \right) = 0.30\)
3. \(\Pr \left( {X > Y} \right) = 0.35\)

Step-by-step solution

Given information

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.

(a) Determine the joint probability function

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

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

(b) Calculate the probability

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

(c) Calculate the probability

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