Posted by **ann** on Thursday, April 14, 2011 at 1:11pm.

A gift store stocks baseball hats in red or green colors. Of the 35 hats on display on a given day, 20 are green. As well, 18 of the hats have a grasshopper logo on the brim. Suppose 11 of the red hats have logos. How many hats are red, or have logos, or both?

The answer in the book is 22. I do not understand how they got this. Could someone please help me? Thanks.

- Math -
**MathMate**, Thursday, April 14, 2011 at 1:41pm
It is not clear if you have learned the inclusion/exclusion principle. If you have, you can apply it without drawing the Venn diagram.

However, for a two-set problem like this, it is easier to draw a Venn diagram and solve accordingly.

There are three sets in a universal set E where the cardinality (i.e. total number of elements) |E|=35.

We are also given that for the set of green hats G, |G|=20.

We conclude therefore that for the set R, |R|=|E|-|G|=35-20=15.

Of the 35 hats, irrespective of colour, 18 of them have logos, so belong to the set L, where |L|=18.

We are required to find the set of hats which are either red, or has a logo, that is, the cardinality of the set R∪L, or |R∪L|.

Consider now the sets L and R.

Draw a Venn diagram for the two, with an intersection, i.e. both red and have a logo. We understand that |L∩R|=11.

So if you put in the Venn diagram 18 for L, 15 for R, and 11 for R∩L. You can calculate that L-R (i.e. with a logo but not red) is 18-11=7, and R-L (i.e. red but no logo) is 15-11=4.

So therefore

|R∪L| = 7+11+4 = 22.

- Math -
**ann**, Thursday, April 14, 2011 at 10:53pm
Thanks a lot.

- Math :) -
**MathMate**, Thursday, April 14, 2011 at 11:22pm
You're welcome!

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