Comments on: Venn Diagrams and Formulas

By: gustavo

gustavo — Mon, 15 Nov 2010 05:43:46 +0000

hola, me podrias todas las formulas por favor, me urge, cuando son excluyentes, binomial etc.
como sacar a unido ab, a interceptado a B, a dado que b.. por favor

By: samjshah

samjshah — Thu, 05 Mar 2009 04:01:18 +0000

@vidyadhar just don’t have the circles intersect/overlap, right?

By: vidyadhar katre

vidyadhar katre — Thu, 05 Mar 2009 03:27:51 +0000

I’m trying to find whether it is possible to represent independent events by venn-diagrams. I have not succeeded so far but would like to receive help from you if you
know how to do it. Thanks.

By: jd2718

jd2718 — Sun, 11 Jan 2009 17:35:29 +0000

Sam,

I find that sometimes are questions are TOO simple…

If the probability was of rolling two dice, and getting, for example, a sum that was odd and less than 3…

Jonathan

By: samjshah

samjshah — Mon, 05 Jan 2009 02:58:12 +0000

@matt: Bah humbug! LAME! You should get it, and write in it when you feel inspired, and when you don’t, let it rot into internet flotsam and jetsam. It doesn’t need to be a thing! You can even write about Joe Millionaire. I promise I’ll read, or at least skim.

By: Matt E

Matt E — Sun, 04 Jan 2009 22:30:46 +0000

Heh, I actually had one for three years, from '01 to '04, though it wasn't specifically about teaching. I've thought about getting back into it, but I feel pressed for time as it is!

By: samjshah

samjshah — Sun, 04 Jan 2009 21:36:21 +0000

@Matt E: when are you going to start a blog?! Think about it!

By: Matt E

Matt E — Sun, 04 Jan 2009 19:41:28 +0000

I know we like to move our students beyond rote formula memorization, but I like the fact that a lot of probability can essentially be boiled down to a couple of formulas, one for “or”, which you mentioned in your post, and one for “and”, which the student did not get quite right.

P(A or B) = P(A) + P(B) – P(A and B)

(If A and B are mutually exclusive, then P(A and B) = 0, making the formula P(A or B) = P(A) + P(B).)

P(A and B) = P(A) P(B|A) = P(B) P(A|B)

(If A and B are independent events, then P(B|A) = P(B) and P(A|B) = P(A), and the formula becomes P(A and B) = P(A) P(B).)

By: samjshah

samjshah — Sun, 04 Jan 2009 16:19:36 +0000

@Dave, I’m not sure that the two unrelated properties is right… because they aren’t unrelated. Maybe because EVERY combination of the two properties is included that it works.

@Matt, I think that’s spot on. I’ll suggest that to the teacher. I guess there is nothing on the surface to suggest they are independent/dependent events. Like when you draw cards/balls without replacement, you KNOW that they are dependent events. But I guess in your head you’re intuitively doing the calculation, or at least KNOW that one will affect the other. You don’t know that for the second question without doing the calculation.

@Kate, totally agreed! I think the teacher assumed that all students would solve it with sample spaces. This kid (precalc) solved it using the formulas, though. Which got us math teachers thinking about how two seemingly similar problems actually had a fundamental difference between them.

@Jackie, yes! I’ll suggest that to the teacher.

Thanks y’all.

By: Jackie

Jackie — Sun, 04 Jan 2009 16:10:03 +0000

I’d ask the student to draw out the sample space to see if that supports his/her calculations.