My Math Forum Probabilty

 October 15th, 2010, 10:46 AM #1 Newbie   Joined: Oct 2010 Posts: 4 Thanks: 0 Probabilty A car dealer has 410 cars on lot. 67 are white, 263 have an auto. transmission, 38 have a sun roof, 34 are white and have an auto. transmission, 11 are white and have a sunroof, 20 have an auto. transmission and a sunroof, and 8 are white, have an auto. transmission and don't have a sunroof. 1. How many are not white, don't have an auto. transmission and don't have a sunroof? 410-263-38-67= 42? 2. How many are white, have an auto. transmission and don't have a sunroof? 34-8=26? These answers seem entirely too simple to me. Any help would be greatly appreciated.
 October 15th, 2010, 11:40 AM #2 Global Moderator     Joined: Nov 2006 From: UTC -5 Posts: 16,046 Thanks: 938 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms Re: Probabilty You subtracted 263 (# with auto. trans.) and 67 (# white) from 410 (the total). But what if there is a car that is both white and has an auto. trans.? You would have subtracted it twice. You need to use inclusion-exclusion.
October 15th, 2010, 03:51 PM   #3
Math Team

Joined: Dec 2006
From: Lexington, MA

Posts: 3,267
Thanks: 408

Re: Probabilty

Hello, carter37!

Is there a typo?
The information is inconsistent .

$\text{Let: }\:\begin{Bmatrix}W=&\text{white} \\ \\ \\ A=&\text{auto.trans.} \\ \\ \\ S=&\text{sunroof} \end{Bmatrix}=$

Quote:
 $\text{A car dealer has 410 cars on lot.}$ $\begin{array}{c}\text{67 are}\,W \\ \\ \\ \text{263 are}\,A \\ \\ \\ \text{38 are}\,S \end{array}\;\;\;\;\; \begin{array}{c}\text{34 are }W\,\text{and}\,A \\ \\ \\ \text{11 are }W\,\text{and}\,S \\ \\ \\ \text{20 are }A\,\text{and}\,S \end{array}\;\;\;\;\; \text{8 are }W\,\text{and}\,A\,\text{and}\,\sim S$

Try to sketch the Venn diagram . . .

$\text{We are told that: }\:n(W \cap\,\! A) \:=\:34$
[color=beige]. . [/color]$\text{and that: }\:n(W \cap\,\! A\,\! \cap \sim S) \:=\:8$

$\text{This means: }\:n(W \cap\,\! A\,\! \cap\,\! S) \:=\:26$

$\text{Then how can }n(W \cap\,\! S) \,=\,11\,?$

$\text{I suspected that the last entry had a typo.}$

$\text{Perhaps it should have been: "8 are }W\,\text{and}\,A\,\text{and}\,S."$

$\text{The Venn diagram could be constructed, but it has }654\text{ cars.}$

 October 15th, 2010, 05:05 PM #4 Newbie   Joined: Oct 2010 Posts: 4 Thanks: 0 Re: Probabilty Soroban: You are absolutely correct, the end should read "8 are white have an automatic transmission and a sunroof." Thanks for pointing that out.
October 16th, 2010, 07:54 AM   #5
Math Team

Joined: Dec 2006
From: Lexington, MA

Posts: 3,267
Thanks: 408

Re: Probabilty

Hello, carter37!

Quote:
 The end should read "8 are white, have an automatic transmission and a sunroof."

Ah, that clears it up!
And I was wrong about the 654 cars.

$(1)\;\;n(W \cup A \cup S) \;=\;n(W)\,+\,n(A)\,+\,n(S)\,-\,n(W\cap A)\,-\,n(A\cap S) \,-\,n(W \cap S) \,+\,n(W\cap A\cap S)$

[color=beige]. . . . . . . . . . . . . [/color]$=\;\;\;67\;\;+\;\;263\;+\;\;38 \;\;-\;\;\;\;\;34 \;\;\;\;\; - \;\;\;\;\; 11 \;\;\;\; -\;\;\;\;20 \;\;\;\;\;\; + \;\;\;\;\;\; 8$

[color=beige]. . . [/color]$n(W \cup A \cup S) \;=\;311$

$\text{Therefore: }\;n(\sim\!W \,\cap\, \sim\! A \,\cap\, \sim\!S) \;=\;410\,-\,311 \;=\;99$

$\text{Your answer to (2) is correct!}$

 Tags probabilty

 Thread Tools Display Modes Linear Mode

 Similar Threads Thread Thread Starter Forum Replies Last Post abhi231594 Algebra 1 January 27th, 2013 09:26 AM percyt Advanced Statistics 3 March 27th, 2012 11:13 PM Student@work Algebra 1 February 12th, 2012 07:06 PM Student@work Algebra 1 February 12th, 2012 06:53 PM finitehelp Advanced Statistics 2 July 31st, 2009 03:33 PM

 Contact - Home - Forums - Cryptocurrency Forum - Top