My Math Forum  

Go Back   My Math Forum > College Math Forum > Advanced Statistics

Advanced Statistics Advanced Probability and Statistics Math Forum


Reply
 
LinkBack Thread Tools Display Modes
April 8th, 2013, 05:09 AM   #1
Member
 
Joined: Jun 2012

Posts: 91
Thanks: 0

Help with answer to odd-numbered exercise

Im currently reading "Introduction to probability" by John E. Freund and I had a hard time understanding one of the exercices in the book and decided to ask you guys for help. The solution is already given, but I still dont get it.

Among the 20 candidates for four positions on a city council, eight are Democrats, eight are Republicans, and four are Independents (that is, they have no official party affiliation).
If the four positions are filled by lot;
(a) What are the odds that two of the chosen candidates will belong to each of the two major parties?

The answer is

784 to 4 061

I suspect this would be quite easy to solve, but still I havent made it, I might have misread the question, or could it be a typo?
Or most probably the reason for this is because Im a newbie in probability maths.
zengjinlian is offline  
 
April 8th, 2013, 12:31 PM   #2
Math Team
 
Joined: Oct 2011
From: Ottawa Ontario, Canada

Posts: 12,748
Thanks: 860

Re: Help with answer to odd-numbered exercise

784 to 4061 is correct.
Your problem can be CLEARLY stated this way:
A box contains 8 black balls, 8 red balls and 4 white balls.
4 balls are drawn at random.
What is probability that at least 1 black ball and at least 1 red ball are picked?

Give it a shot!!
Denis is offline  
April 8th, 2013, 10:02 PM   #3
Member
 
Joined: Jun 2012

Posts: 91
Thanks: 0

Re: Help with answer to odd-numbered exercise

Thanks for your reply Denis! I actually got it now, realizing Ive misread the question..

Quote:
Originally Posted by Denis
What is probability that at least 1 black ball and at least 1 red ball are picked?

Okey, the following scenarios counts as success according to your criteria

3 black + 1 red = (8,3)*(8,1) = 448 combos
2 black + 2 red = (8,2)*(8,2) = 784 combos
2 black + 1 red = (8,2)*(8,1)= 224 combos
1 black + 3 red = (8,1)*(8,3)=448 combos
1 black + 2 red = (8,2)*(8,1)= 224 combos
1 black + 1 red = (8,1)*(8,1)= 64 combos

So to answer your question is 2192/4845,( the odds are 2192 to 2653)
The right answer to the original question is marked in bold letters.

It turs out I misread the question , as far as I was understanding the author was looking for the odds that EXACTLY two of the four chosen candidates is a Democrat and a Republican thus leaving two positions for Independents.
But instead he was looking for the odds that exactly two Democrats and exactly two Republicans where choosen, leaving no position for Independents.
Glad I finally solved it.
zengjinlian is offline  
April 9th, 2013, 07:38 AM   #4
Newbie
 
Joined: Mar 2013

Posts: 3
Thanks: 0

Re: Help with answer to odd-numbered exercise

You want :

2 Republicans among 8,
2 Democrats among 8,
0 Independant among 4.

It makes :



It is the number of possibilities that respect your goal.

You choose :

4 senators among 20 candidates.

It makes :



I may be wrong somewhere.
frezard is offline  
April 9th, 2013, 11:34 AM   #5
Member
 
Joined: Jun 2012

Posts: 91
Thanks: 0

Re: Help with answer to odd-numbered exercise

I forgot to include the white balls...

here is my edited calculation for Denis scenario, to get at least 1 black and 1 red.




Quote:
Originally Posted by zengjinlian
3 black + 1 red + 0 white = (8,3)*(8,1) = 448 combos
2 black + 2 red + 0 white = (8,2)*(8,2) = 784 combos (J.E Freunds case)
2 black + 1 red + 1 white = (8,2)*(8,1)*(4,1) = 896 combos
1 black + 3 red + 0 white = (8,1)*(8,3)=448 combos
1 black + 2 red + 1 white= (8,2)*(8,1)*(4,1)= 896 combos
1 black + 1 red + 2 white = (8,1)*(8,1)*(4,2)= 384 combos
Shows that in 3856 different combinations of the possible 4845 combinations we will have at least 1 black and 1 red ball (or 1 Democrat and 1 Republican)

So the odds are 3856 to 989 for this scenario. And the probability is 0.79

By the way Frezzard did a correct explanation of the scenario as presented by J.E Freund (even if he left some room for misreading the question)
zengjinlian is offline  
April 9th, 2013, 08:06 PM   #6
Math Team
 
Joined: Oct 2011
From: Ottawa Ontario, Canada

Posts: 12,748
Thanks: 860

Re: Help with answer to odd-numbered exercise

Quote:
Originally Posted by zengjinlian
Shows that in 3856 different combinations of the possible 4845 combinations we will have at least 1 black and 1 red ball (or 1 Democrat and 1 Republican)

So the odds are 3856 to 989 for this scenario. And the probability is 0.79
Correct.
Denis is offline  
Reply

  My Math Forum > College Math Forum > Advanced Statistics

Tags
answer, exercise, oddnumbered



Thread Tools
Display Modes


Similar Threads
Thread Thread Starter Forum Replies Last Post
Can anyone help me with this exercise? alfred_oh Advanced Statistics 1 April 30th, 2013 08:35 AM
numeric entry problem with answer..but I think the answer is rage Algebra 2 September 14th, 2012 09:27 PM
brief exercise icemanfan Number Theory 2 March 15th, 2012 04:58 PM
No. of ways to seat round a table (numbered seats) Punch Algebra 4 February 1st, 2012 10:34 PM
Could someone help me with this exercise Touya Akira Abstract Algebra 8 May 10th, 2011 07:57 AM





Copyright © 2018 My Math Forum. All rights reserved.