Chapter 6: Q.6.11 (page 275)

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

Expert verified

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

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,

e^{- x} > 0, \forall x \in (0, \infty) a n d e^{- 2 y} > 0, \forall y \in (0, \infty) t h e r e f o r e t r i v i a l l y f (x, y) = e^{- x} e^{- 2 y} f (x, y) \geq 0

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

$\iint f (x, y) d x d y = \int_{- \infty}^{\infty} \int_{0}^{\infty} 2 e^{- x} e^{- 2 y} d x d y \iint f (x, y) d x d y = 2 \int_{- \infty}^{\infty} e^{- 2 y} d y \int_{0}^{\infty} 2 e^{- x} d x ∭_{- \infty} f (x, y) d x d y \int {[\frac{e^{- 2 y}}{- 2}]}_{- \infty}^{\infty} {[\frac{e^{- x}}{- 1}]}_{- \infty}^{\infty} \int e^{a x} d x = \frac{e^{x}}{a}] \iint f (x, y) d x d y = 2 [\frac{e^{- \infty}}{- 2} - \frac{e^{- \infty}}{- 2}] [\frac{e^{- \infty}}{- 1} - \frac{e^{- x}}{- 1}] \iint f (x, y) d x d y = 1$

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

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