Question

A television store owner figures that 45 percent of the customers entering his store will purchase an ordinary television set, 15 percent will purchase a plasma television set, and 40 percent will just be browsing. If 5 customers enter his store on a given day, what is the probability that he will sell exactly 2 ordinary sets and 1 plasma set on that day?

Short answer

The joint density of$f(x,y)=2e^{-x}{e}^{-2y};\forall 0<x,0<y<\infty$

Step-by-step solution

Content Introduction

A television store owner estimates that 45 percent of clients entering his business will buy a regular television set, 15% will buy a plasma television set, and 40% will simply browse. On any given day, he will have 5 clients at his store.

Content Explanation

We are given,

$$\begin{gathered} e^{-x}>0,\forall x\in (0,\infty )and{e}^{-2y}>0,\forall y\in (0,\infty )thereforetriviallyf(x,y)=e^{-x}{e}^{-2y} \\ f(x,y)\ge 0 \end{gathered}$$

Hence (A) holds for the given bivariate function. Now check for B:

$$\begin{gathered} \iint f(x,y)dxdy={\int }_{-\infty }^{\infty }{\int }_0^{\infty }2e^{-x}{e}^{-2y}dxdy \\ \iint f(x,y)dxdy=2{\int }_{-\infty }^{\infty }{e}^{-2y}dy{\int }_0^{\infty }2e^{-x}dx \\ {\iiint }_{-\infty }f(x,y)dxdy \\ \left.\int {\left[\frac{e^{-2y}}{-2}\right]}_{-\infty }^{\infty }{\left[\frac{e^{-x}}{-1}\right]}_{-\infty }^{\infty }\int e^{ax}dx=\frac{e^x}{a}\right] \\ \iint f(x,y)dxdy=2\left[\frac{e^{-\infty }}{-2}-\frac{e^{-\infty }}{-2}\right]\left[\frac{e^{-\infty }}{-1}-\frac{e^{-x}}{-1}\right] \\ \iint f(x,y)dxdy=1 \end{gathered}$$

Hence, B holds for the given bivariate function f (x , y) . Therefore, $f(x,y)=2e^{-x}{e}^{-2y};\forall 0<x,0<y<\infty$is a joint density of ( x , y).