## Friday, March 31, 2006

### A wedding feast

At the Miller's wedding, it was determined that 90% of the guests had chicken, 80% had peas, 70% had vanilla cake, and 60% had ice cream. No one, however, had all four items. What percentage of the guests had at least one of the two deserts?

1. I love these. Clever idea for a blog. I'll be back to find the answer. I think I know, but not ready to commit.

2. Jan, I'm glad you like it. Please come back and try themout!

90% of the guests had chicken, 80% had peas, 70% had vanilla cake, and 60% had ice cream. No one, however, had all four items. What percentage of the guests had at least one of the two deserts?

Answer: Let C be the set of guests who had chicken, P for Peas, V for Vanilla cake and I for ice cream. Then let C' represent those who didn't have chicken and so forth. Then C' = 10%, P' = 20%, V' = 30% and I' = 40%. Because no one had all four items, the union of this set is the set of all guests. And since 10+20+30+40 = 100%, these four sets are disjoint. In particular, that means that V' and I' have no members in common, so everyone must have had at least one of the two deserts.

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