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Domenico writes: | Domenico writes: | ||
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What intrigued me was the fact that P(C1=T | C2=T) = P(C2=T | C1=T), and this holds regardless of the numbers of the two types of cats. Here: | |||
:C1 = the type of the first cat that comes out of the bag | :C1 = the type of the first cat that comes out of the bag | ||
:C2 = the type of second cat that comes out of the bag | :C2 = the type of second cat that comes out of the bag | ||
:T = tabby | :T = tabby | ||
</blockquote> | |||
Revision as of 21:17, 22 August 2013
Marilyn on letting-cats-out-of-the-bag
Domenico Rosa sent a link to the following:
Letting the cat out of the bag: Can you solve these 2 puzzles?
by Marilyn vos Savant, Parade, 10 August 2013
A reader poses the following question:
Say you have a bag full of cats. Two are tabbies, and three are calicos. You let one cat out of the bag. It runs up a tree before you get a chance to see its color. Then you let out another cat. As you pry its jaws from your ankle, you see that it’s a tabby. What are the chances that the cat in the tree is also a tabby?
Domenico writes:
What intrigued me was the fact that P(C1=T | C2=T) = P(C2=T | C1=T), and this holds regardless of the numbers of the two types of cats. Here:
- C1 = the type of the first cat that comes out of the bag
- C2 = the type of second cat that comes out of the bag
- T = tabby