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 Posted: Mon, 5 Mar 2012 00:20:07 UTC
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Joined: Mon, 28 Nov 2011 01:12:03 UTC
Posts: 10
This is the original problem that my question is coming out of:

A number X is chosen at random from the series 2,5,8,... and another number Y is chosen at random from the series 3, 7, 11, ... Each series has 100 terms. Find P[ X=Y ].

The answer to this problem is to see that 100^2 = 10,000 equally likely possible choices of (X,Y). Of these choices, the pairs that equal X and Y are (11,11), (23,23), (35,35), (299,299), etc. There are 25 such pairs, so the probability is 25/10,000.

My question is what is the proper mathematical way to determine those pairs of (X = Y)? The answer key tells you that they are of the form (12k - 1, 12k - 1). Working backwards from that, I see that since the formulas for the two sequences are:

2 + 3(n - 1),

3 + 4(n - 1),

respectively, you can multiply each out to get:

3n - 1,

4n - 1.

Is finding the common terms as simple as just multiplying the coefficients of n for each? Would it matter if their constants weren't equal, unlike in this case?

This isn't touched on directly in the book (it's for the Actuarial P/1 Exam), and I haven't been able to find anything on the internet about finding common terms of sequences.

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 Posted: Mon, 5 Mar 2012 01:01:02 UTC

Joined: Sat, 26 Apr 2003 22:14:40 UTC
Posts: 2167
Location: El Paso TX (USA)
Just staring at the two sequences (sic!), one can see that exactly every fourth term of the first sequence will have a match in the second sequence. 100/4=25.

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 Posted: Mon, 5 Mar 2012 01:20:41 UTC
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Joined: Mon, 28 Nov 2011 01:12:03 UTC
Posts: 10
helmut wrote:
Just staring at the two sequences (sic!), one can see that exactly every fourth term of the first sequence will have a match in the second sequence. 100/4=25.

But it can't be a matter of just eyeballing the sequence though, right? (though if that's what it takes, I'll do that.) I just have no idea and have been unable to identify a standard mathematical way to do it.

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 Posted: Mon, 5 Mar 2012 03:04:22 UTC
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Joined: Wed, 30 Mar 2005 04:25:14 UTC
Posts: 15562
Location: Austin, TX
Teknontheou wrote:
helmut wrote:
Just staring at the two sequences (sic!), one can see that exactly every fourth term of the first sequence will have a match in the second sequence. 100/4=25.

But it can't be a matter of just eyeballing the sequence though, right? (though if that's what it takes, I'll do that.) I just have no idea and have been unable to identify a standard mathematical way to do it.

Eyeballing is fine since you know the sequence is arithmetic. "Eyeballing" means "this is so standard everyone remembers how to prove it if they have to, but it would be a waste of time to really do". Helmut did the eyeballing, but he wasn't suggesting you write that as your answer, he means to tell you what's going on colloquially so you can formalize it into math for your final solution.

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 Posted: Mon, 5 Mar 2012 06:45:34 UTC
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Joined: Sun, 24 Jul 2005 20:12:39 UTC
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Location: Ottawa Ontario
Hmmm...seems to me the major problem is the 1st term of each sequence;
Seq#1 has 1st term = 2: 2 5 8 11 14 ..... 293 296 299 END
Seq#2 has 1st term = 292: 292 296 300 ......
So here there's only 1 possibility.

Why not force the 1st terms to equal the common difference;
3 6 9
4 8 12
or something like 1st term = common difference - 1

The easy enough to work out a general case formula.

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