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Chapter 5: Q8E (page 211)

A stockroom currently has \({\rm{30}}\) components of a certain type, of which \({\rm{8}}\) were provided by supplier \({\rm{1,10}}\)by supplier \({\rm{2}}\) , and \({\rm{12}}\) by supplier \({\rm{3}}\). Six of these are to be randomly selected for a particular assembly. Let \({\rm{X = }}\) the number of supplier l's components selected, \({\rm{Y = }}\) the number of supplier \({\rm{2}}\) 's components selected, and \({\rm{p(x,y)}}\) denote the joint pmf of \({\rm{X}}\) and\({\rm{Y}}\).

a. What is \({\rm{p(3,2)}}\) ? (Hint: Each sample of size \({\rm{6}}\) is equally likely to be selected. Therefore, \({\rm{p(3,2) = }}\) (number of outcomes with \({\rm{X = 3}}\) and \({\rm{Y = 2)/(}}\) total number of outcomes). Now use the product rule for counting to obtain the numerator and denominator.)

b. Using the logic of part (a), obtain \({\rm{p(x,y}}\) ). (This can be thought of as a multivariate hypergeometric distribution-sampling without replacement from a finite population consisting of more than two categories.)

Answer

Short Answer

Expert verified

a) The value of \({\rm{p}}\left( {{\rm{3,2}}} \right)\) is \({\rm{p}}\left( {{\rm{3,2}}} \right){\rm{ = 0}}{\rm{.0509}}\).

b) The solution is \({\rm{p(x,y) = }}\left\{ {\begin{array}{*{20}{l}}{\frac{{\left( {\begin{aligned}{*{20}{c}}{\rm{8}}\\{\rm{x}}\end{aligned}} \right)\left( {\begin{aligned}{*{20}{c}}{{\rm{10}}}\\{\rm{y}}\end{aligned}} \right)\left( {\begin{aligned}{*{20}{c}}{{\rm{12}}}\\{{\rm{6 - x - y}}}\end{aligned}} \right)}}{{\left( {\begin{aligned}{*{20}{c}}{{\rm{30}}}\\{\rm{6}}\end{aligned}} \right)}}}&{{\rm{,0~x + y~6 and x,y^I }}{{\rm{N}}_{\rm{0}}}}\\{\rm{0}}&{{\rm{, otherwise}}{\rm{. }}}\end{array}} \right.\)

Step by step solution

01

Definition

Probability simply refers to the likelihood of something occurring. We may talk about the probabilities of particular outcomes鈥攈ow likely they are鈥攚hen we're unclear about the result of an event. Statistics is the study of occurrences guided by probability.

02

Given in question

There are total of \({\rm{30}}\) components

\({\rm{8}}\)provided by supplier \({\rm{1}}\),

\({\rm{10}}\)provided by supplier \({\rm{2}}\),

\({\rm{12}}\)provided by supplier \({\rm{3}}\).

Six of these are randomly selected.

03

Determining the value of \({\rm{p}}\left( {{\rm{3,2}}} \right)\)

a)

As explained in the hint, we have that

\({\rm{p(3,2) = }}\frac{{\left( {\begin{array}{*{20}{l}}{\rm{8}}\\{\rm{3}}\end{array}} \right)\left( {\begin{array}{*{20}{c}}{{\rm{10}}}\\{\rm{2}}\end{array}} \right)\left( {\begin{aligned}{*{20}{c}}{{\rm{12}}}\\{\rm{1}}\end{aligned}} \right)}}{{\left( {\begin{aligned}{*{20}{c}}{{\rm{30}}}\\{\rm{6}}\end{aligned}} \right)}}\) Where \(\left( {\begin{aligned}{*{20}{l}}{\rm{8}}\\{\rm{3}}\end{aligned}} \right)\)

represents the number of ways to take \({\rm{3}}\) components from supplier\({\rm{1}}\),

\(\left( {\begin{aligned}{*{20}{c}}{{\rm{10}}}\\{\rm{2}}\end{aligned}} \right)\)

represents the number of ways to take \({\rm{2}}\) components from supplier \({\rm{2}}\),

\(\left( {\begin{aligned}{*{20}{c}}{{\rm{12}}}\\{\rm{1}}\end{aligned}} \right)\)

represents the number of ways to take \({\rm{1}}\) component from the supplier \({\rm{3}}\). By the product rule for counting, the numerator would be

\(\left( {\begin{aligned}{*{20}{l}}{\rm{8}}\\{\rm{3}}\end{aligned}} \right)\left( {\begin{aligned}{*{20}{c}}{{\rm{10}}}\\{\rm{2}}\end{aligned}} \right)\left( {\begin{aligned}{*{20}{c}}{{\rm{12}}}\\{\rm{1}}\end{aligned}} \right){\rm{ = 56 \times 45 \times 12 = 30,240}}\)

The denominator is total number of outcomes, which is the number of ways to take \({\rm{6}}\) components out of\({\rm{30}}\), hence

\(\left( {\begin{aligned}{*{20}{c}}{{\rm{30}}}\\{\rm{6}}\end{aligned}} \right){\rm{ = 593,775}}\)

Now, the requested probability is

\({\rm{p(3,2) = }}\frac{{{\rm{30,240}}}}{{{\rm{593,775}}}}{\rm{ = 0}}{\rm{.0509}}\)

04

Obtaining the value of \({\rm{p}}\left( {{\rm{x,y}}} \right)\)

(b):

The bivariate (multivariate) hypergeometric distribution, with the same logic as above, is given with

\({\rm{p(x,y) = }}\left\{ {\begin{aligned}{*{20}{l}}{\frac{{\left( {\begin{aligned}{*{20}{c}}{\rm{8}}\\{\rm{x}}\end{aligned}} \right)\left( {\begin{aligned}{*{20}{c}}{{\rm{10}}}\\{\rm{y}}\end{aligned}} \right)\left( {\begin{aligned}{*{20}{c}}{{\rm{12}}}\\{{\rm{6 - x - y}}}\end{aligned}} \right)}}{{\left( {\begin{aligned}{*{20}{c}}{{\rm{30}}}\\{\rm{6}}\end{aligned}} \right)}}}&{{\rm{,0~x + y~6 and x,y^I }}{{\rm{N}}_{\rm{0}}}}\\{\rm{0}}&{{\rm{, otherwise}}{\rm{. }}}\end{aligned}} \right.\)

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