simple math question.

[gotta240]

Been way too long since my last math class.


Probability of event A happening is 2 percent.

Probability of event B happening is 4 percent.

What is the probability that BOTH will happen at the same time.


I THINK its 8/10000 which is .0008%

Is that correct?

[Sil-Abc]

both of them multiplied together so .02x.04

[stinky_180]

.02 *.04 = .0008

Yesss....?

[eastcoastS14]

both of them multiplied together so .02x.04

I dont think this is right...

edit: maybe it is...im terrible at math

[just1pepsi]

Im too dumb for this, but I can use google quite well. read this, maybe it will help with your simple math problem:

http://richardbowles.tripod.com/maths/probability/prob.htm

[Xandy Boosts]

Yes I think you guys are right

What you're asking for is the P(A upside down U B) or P(A and B)

If A and B are Independent
A and B are two events. If A and B are independent, then the probability that events A and B both occur is:

p(A and B) = p(A) x p(B).

[slider2828]

Yes if it is P and B it is multiplied.

[gotta240]

lol. So is my answer right or wrong...that simple guys!

[exitspeed]

The answer is in your rep...

[ALTRNTV]

p(A and B) = p(A) x p(B)

then you have to get the
tangent of (A and B), then
the square root the logarithm,
then solve for P

I hate math. :bigok:

[MikeisNissan]

I h8 Math!!!!!!!!

[gabe30]

p(A and B) = p(A) x p(B)

then you have to get the
tangent of (A and B), then
the square root the logarithm,
then solve for P

I hate math. :bigok:

lol even more so complicated.

stats: if P(A) and P(B) are independent (this is important that they are or are not) then your answer is P(A and B) = P(A) x P(B) = .02 x .04 = .0008

[Antihero983]

I have to ask, what the fuck kind of math is this?!

I was an english major for a reason.....

[gabe30]

this is familiar to me because im taking intro to statistics

[gotta240]

i'm using it to determine the probability that a random infant has a disease.

The percentage of adults with the disease is 2 percent. The chance that a mother with the disease will pass it to her child is 4 percent. Therefore, the chance that a random infant will have the disease (assuming she can only get it from her mother) is EXTREMELY low.

Something like a 1 in 1250 chance, aka .0008. Correct?

[BC240]

Yeah that's right

[iwishiwas-all*]

AHHH stats all over again... but its P(a U b) so .02 x .04

[Andrew Bohan]

i minored in math

:eek3:

[ATLspeed]

its 22 becuz 2+2=22................rriiiigggghhhttt...

[blackrms13]

if its AND u multiply
if its OR u add

im in stat100...hahaaa

[YoungGun]

Math is the Devil.

[stinky_180]

lol so what is the answer?

[Andrew Bohan]

2% = .02
4% = .04
.02 x .04 = .0008 = .08%