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'''Discussion'''<br> | '''Discussion'''<br> | ||
In general, under what conditions on events A and B does P(A | B) = P(B | A)? Do you see why these conditions hold above (regardless of the number of the two types)? |
Revision as of 21:29, 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
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?
You can read Marilyn's answer at the link above, but of course you should first try to solve the puzzle on your own!
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
Discussion
In general, under what conditions on events A and B does P(A | B) = P(B | A)? Do you see why these conditions hold above (regardless of the number of the two types)?