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 May 16th, 2009, 08:01 PM #1 Newbie   Joined: May 2009 Posts: 6 Thanks: 0 Help with probability problem. So I saw this math problem on an internet forum a while ago and I decided to take a shot at it. The first way I tried it gave me a wrong answer. The second way gave me the correct answer. I've been trying to figure out why the first attempt was wrong when, logically, it should be equal. The problem went something like this. "You have a population of men and women. The chance of picking a single man is 1/3. If picking 2 men(without replacement) yields a probability of 1/10, how many people are in the population?" On the left is the correct way(my second attempt) to do it but the right is identical, only using decimals instead of fractions. So any idea why it doesn't work? Also, the bottom portion is a random attempt that got me another weird answer... Seriously, I've been racking my brain on this and just can't get it. My stats teacher was stumped as well so if anyone can help me out I would love you forever. =)
 May 17th, 2009, 05:21 PM #2 Global Moderator   Joined: May 2007 Posts: 6,665 Thanks: 651 Re: Help with probability problem. The attachement is unreadable.
 May 17th, 2009, 07:03 PM #3 Senior Member   Joined: Jul 2008 Posts: 895 Thanks: 0 Re: Help with probability problem. How do you, for example, get .3(repeated]x.3(repeated) = .1(repeated)? Also, you can not divide out added terms from the numerator and denominator. The right side is just plain rubbish. You need to follow the rules of arithmetic and algebra.
May 18th, 2009, 06:20 AM   #4
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Re: Help with probability problem.

Quote:
 Originally Posted by Dave How do you, for example, get .3(repeated]x.3(repeated) = .1(repeated)? Also, you can not divide out added terms from the numerator and denominator. The right side is just plain rubbish. You need to follow the rules of arithmetic and algebra.
ummm .3 repeated x .3 repeated is equal to .1 repeated. (1/3)(1/3)=(1/9)=.1 repeated. And can you point out exactly where I didn't follow the correct rules? It seems people can only criticize and not actually help

 May 18th, 2009, 07:26 AM #5 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,900 Thanks: 1094 Math Focus: Elementary mathematics and beyond Re: Help with probability problem. $\frac{.\bar{3}n}{n}$ is equivalent to $\frac{n}{3n}$ but $\frac{.\bar{3}n-1}{n-1}$ is not equivalent to $\frac{n-1}{3n-1}$ .
May 19th, 2009, 09:11 PM   #6
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Re: Help with probability problem.

Quote:
 Originally Posted by greg1313 $\frac{.\bar{3}n}{n}$ is equivalent to $\frac{n}{3n}$ but $\frac{.\bar{3}n-1}{n-1}$ is not equivalent to $\frac{n-1}{3n-1}$ .
I appreciate the help but would you mind explaining why they aren't equal? Or how to convert it to a proper form. Is it simply impossible to compute in decimal form?

May 20th, 2009, 04:57 PM   #7
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Re: Help with probability problem.

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Originally Posted by ListerineStrips
Quote:
 Originally Posted by greg1313 $\frac{.\bar{3}n}{n}$ is equivalent to $\frac{n}{3n}$ but $\frac{.\bar{3}n-1}{n-1}$ is not equivalent to $\frac{n-1}{3n-1}$ .
I appreciate the help but would you mind explaining why they aren't equal? Or how to convert it to a proper form. Is it simply impossible to compute in decimal form?
$\frac{.\bar{3}n}{n}$ = 1/3. Just do the arithmetic.

May 20th, 2009, 05:28 PM   #8
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Re: Help with probability problem.

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 Originally Posted by ListerineStrips I appreciate the help but would you mind explaining why they aren't equal? Or how to convert it to a proper form. Is it simply impossible to compute in decimal form?
There is absolutely no point converting something to decimals unless there is a valid reason to do so. For instance, if you are asked to solve the equation $x^2-4x-4=0,$ the solutions are exactly $x=2+2\sqrt{2}$ and $x=2-2\sqrt{2}.$ Writing them as $x=4.828$ and $x=-0.828$ is not only pointless, it is also wrong. In general, only use decimals if you are given decimals as input or if you are dealing with physical quantities.

In your problem, there is no need to convert to decimals - if you must, though, note that

$\frac{n-1}{3n-1}=\frac{n/3-1/3}{n-1/3}=\frac{.\dot3n-.\dot3}{n-.\dot3}.$

This should give the correct answer.

May 20th, 2009, 07:14 PM   #9
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Re: Help with probability problem.

Quote:
Originally Posted by mattpi
Quote:
 Originally Posted by ListerineStrips I appreciate the help but would you mind explaining why they aren't equal? Or how to convert it to a proper form. Is it simply impossible to compute in decimal form?
There is absolutely no point converting something to decimals unless there is a valid reason to do so. For instance, if you are asked to solve the equation $x^2-4x-4=0,$ the solutions are exactly $x=2+2\sqrt{2}$ and $x=2-2\sqrt{2}.$ Writing them as $x=4.828$ and $x=-0.828$ is not only pointless, it is also wrong. In general, only use decimals if you are given decimals as input or if you are dealing with physical quantities.

In your problem, there is no need to convert to decimals - if you must, though, note that

$\frac{n-1}{3n-1}=\frac{n/3-1/3}{n-1/3}=\frac{.\dot3n-.\dot3}{n-.\dot3}.$

This should give the correct answer.
Thank you! This was an extremely helpful response and I seriously appreciate it. You're right that it would be foolish to calculate anything with decimals when it's not needed, I just wanted to do it to satisfy my own curiosity. Again, thank you very much!

May 20th, 2009, 07:40 PM   #10
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Re: Help with probability problem.

Quote:
 Originally Posted by ListerineStrips It seems people can only criticize and not actually help
Perhaps you should browse through this site and count the number of times I have helped, and the number of times I have criticised. Then count the number of times you have helped and the number of times you have criticised [Count this as 1.] Mathman said it was unreadable, and didn't bother any further. I should have followed his example. However, I tried to read and interpret what you had done. That was being helpful, not critical. Further, any help I give here is done under adverse circumstances, and need not be done at all in fact. I'll admit my wording was not too clear, but I was trying to interpret what you might have been thinking, and it did seem that you were cancelling where you shouldn't, which was the real point I was making.

For clarity in future, I'd recommend downloading and using MathType to write your formulas, showing clear steps to leave no doubt. You'll have to learn how to use it, but it does make a big difference.

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