My Math Forum Help with answer to odd-numbered exercise

 April 8th, 2013, 06:09 AM #1 Member   Joined: Jun 2012 Posts: 91 Thanks: 0 Help with answer to odd-numbered exercise I´m 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 haven´t made it, I might have misread the question, or could it be a typo? Or most probably the reason for this is because I´m a newbie in probability maths.
 April 8th, 2013, 01:31 PM #2 Math Team   Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 13,986 Thanks: 995 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!!
April 8th, 2013, 11:02 PM   #3
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Re: Help with answer to odd-numbered exercise

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.

 April 9th, 2013, 08:38 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 : $\frac{8!}{2!(8-2)!}\frac{8!}{2!(8-2)!}\frac{4!}{0!(4-0)!}= 784$ It is the number of possibilities that respect your goal. You choose : 4 senators among 20 candidates. It makes : $\frac{20!}{4!(20-4)!}= 4845$ I may be wrong somewhere.
April 9th, 2013, 12:34 PM   #5
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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)

April 9th, 2013, 09:06 PM   #6
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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.

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