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This problem is from Problem Solving Strategies - Crossing the River with Dogs and Other Mathematical Adventures by Ken Johnson and Ted Herr.

I first let capital letters denote shirts, and lowercase letters denote ties (with one letter representing each to make things easier).

So, B = blue, G = green, W = white, P = blue and green print, and S = red and white striped shirt.
I also made w = white, b = blue, y = yellow, p = green and blue striped tie

Then, I began working out the possible combinations:

For (color) shirt:

Blue (no blue tie, pattern much include blue [as denoted in the problem]): Bw, By, Bp

Green (no green tie, patter much include green): Gw, Gb, Gy, Gp

White: Wb, Wy

Blue and Green Pattern (P): Pb

Red and white stripe (S): Sw

I am wondering if there are any possibilities I missed. As of now, I have worked out there to be 11 combinations, with 6 out of the 11 involving the color blue. Any help would be much appreciated.

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Electronics: Method OK, for the patterns I would have tried to use names that reminded me of colours. Like BG for blue-green shirt. My count of the combinations that have some blue is $7$. –  André Nicolas Mar 3 '12 at 1:20
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I only seem to have six on my list: Bw, By, Bp, Gb, Wb, Pb Which one do you have that I am missing? Do you have more than 11 combinations (with 6/11 having some blue)? –  Joe Mar 3 '12 at 1:23
    
There's a lot of information about which combinations Rudy considers valid, but none at all about how he chooses between the valid ones. Is there an implicit assumption that he will pick every valid combination with the same probability? (Rather than, say, pick a random shirt with equal distribution and then pick the first tie that can go with it -- among many other possible strategies). –  Henning Makholm Mar 3 '12 at 1:25
    
Edit: The problem stated above is verbatim from the text book. I think it is safe to assume equal distribution when it comes to picking the shirts and ties, as this is a basic problem solving book for the most part. –  Joe Mar 3 '12 at 1:28
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@Jay Electronics: Your list was complete, and I don't know which you forgot to count for the blues part. Probably one of the p or P ones. That's partly because p (and P) is poor notation, it doesn't scream blue so is easy to overlook. –  André Nicolas Mar 3 '12 at 1:37

1 Answer 1

up vote 2 down vote accepted

One relatively easy systematic approach is to make a table for the combinations; here I’ve listed the shirts down the side and the ties across the top. I’ve put a $b$ in the cells corresponding to acceptable combinations that include blue, and a $y$ in the cells corresponding to other acceptable combinations.

$$\begin{array}{r|c} &\text{white}&\text{blue}&\text{yellow}&\text{green/blue}\\ \hline \text{blue}&b&&b&b\\ \hline \text{green}&y&b&y&b\\ \hline \text{white}&&b&y\\ \hline \text{blue/green}&&b\\ \hline \text{red/white}&y \end{array}$$

As you can see, there are indeed $11$ acceptable combinations, but seven contain blue. Thus, if he is equally likely to pick any of the acceptable combinations, his probability of picking one that includes blue is $\frac7{11}$.

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Ah. Clever solution with the table - nice way to organize the information! Thanks. –  Joe Mar 3 '12 at 1:36

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