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November 4

Conditional probability question

Hi. I'm having a difficult time with some probability notation. I have some probability events A and B that satisfy P(B) = P(B|A)P(A). But P(A) itself is a conditional probability P(C|D). How do I write this out properly in terms of B, C and D? P(B) = P(B|C,D) P(C|D) ?? Many thanks. --18.189.121.106 (talk) 09:25, 4 November 2011 (UTC)[reply]

I don't think this makes sense. You can condition a conditional probability but not on a conditional probability. P((B|C)|D) makes sense and is equal to P(B|C,D), but I know of no interpretation for P(B|(C|D)). C|D is not an event. -- Meni Rosenfeld (talk) 11:22, 4 November 2011 (UTC)[reply]

I omit the P and write just (A|B) for the probability of A conditioned on B. This is not standard, however. The unconditional probability of A is then written (A| ). The rule for conditional probability is (A,B|)=(A|B)(B|). Actually I identify the events by numbers rather than by letters. A=10. B=01. (A AND B)=11. (NOT A)=20. (NOT B)=02. Then the rule for conditional probability is written (11|00)=(10|01)(01|00). Bo Jacoby (talk) 12:53, 4 November 2011 (UTC).[reply]

i concur — Preceding unsigned comment added by 203.112.82.1 (talk) 19:51, 6 November 2011 (UTC)[reply]

graphs of functions

how to draw graph of f(x,y) in general. — Preceding unsigned comment added by nkp.59.165.108.89 (talk) 12:49, 4 November 2011 (UTC)[reply]

It cannot be done. --COVIZAPIBETEFOKY (talk) 12:54, 4 November 2011 (UTC)[reply]

For well-behaved functions

\mathbb {R} \times \mathbb {R} \rightarrow \mathbb {R}

surely it's possible to approximate the ideal graph by sampling and plotting that with a program? « Aaron Rotenberg « Talk « 13:33, 4 November 2011 (UTC)[reply]

Long ago Math geeks stole Contour lines to show how continuous functions of z = f(x,y) worked, but that's fallen down a black hole as of late. Hcobb (talk) 14:20, 4 November 2011 (UTC)[reply]

Plot (graphics) might be worth a look to see what kind of thing you want. Dmcq (talk) 14:36, 4 November 2011 (UTC)[reply]

In general it's not possible, but take a look at our article on the implicit function theorem. — Fly by Night (talk) 13:39, 5 November 2011 (UTC)[reply]

thanks to all of u.