anonymous
  • anonymous
HELP PLEASE!! Calculate probability of the independent event P(A or B) when P(a)=.3 and P(B)=.7
Mathematics
chestercat
  • chestercat
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anonymous
  • anonymous
You just have to multiply .3 by .7
anonymous
  • anonymous
oh ok well what about conditional probability with the same variables?
ybarrap
  • ybarrap
$$ P(A\cup B)=P(A)+P(B)-P(A\cap B) $$ Since A and B are independent, \(P(A\cap B)=0\) So just SUM the independent probabilities.

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anonymous
  • anonymous
so just 1?
ybarrap
  • ybarrap
Yes. See - http://en.wikipedia.org/wiki/Inclusion%E2%80%93exclusion_principle
ybarrap
  • ybarrap
Specifically, this section - http://en.wikipedia.org/wiki/Inclusion%E2%80%93exclusion_principle#In_probability
anonymous
  • anonymous
ok what about if the variables are P(a)=.5 P(b)=.4 and then P(a and b)=.1?????
anonymous
  • anonymous
I'm supposed to calculate the probability
ybarrap
  • ybarrap
You've got everything you need, P(a), P(b) and P(a and b) -- see my formula above! Just plug in for P(a or b).
ybarrap
  • ybarrap
$$ P(A\cup B)=P(A)+P(B)-P(A\cap B)=0.5+0.4-0.1 $$ See how I did that?
anonymous
  • anonymous
yes i just thought a and b were supposed to = P(A and B) but the answer is .8
ybarrap
  • ybarrap
Are you asking me?
anonymous
  • anonymous
yes(:
ybarrap
  • ybarrap
What are you asking, not sure
anonymous
  • anonymous
the answer is .8 right?
ybarrap
  • ybarrap
Yes for P(a or b).
anonymous
  • anonymous
ok but theres nothing else i need to find? it just says calculate the probability
ybarrap
  • ybarrap
Tell me the whole question to be sure
anonymous
  • anonymous
ybarrap
  • ybarrap
Oh, those are different problems that P(A or B) Ok
ybarrap
  • ybarrap
Do you know what P(A|B) means?
anonymous
  • anonymous
i know its a conditional probability problem
ybarrap
  • ybarrap
$$ P(A|B)=\cfrac{PA\cap B}{P(B)}=\cfrac{0.1}{0.4} $$ You can do the next. BRB
anonymous
  • anonymous
i got .25
ybarrap
  • ybarrap
Do #27 now
anonymous
  • anonymous
umm isn't that what i did?
anonymous
  • anonymous
oh okay for 27 i got 4
anonymous
  • anonymous
if you can check the other ones i got .21 on 23 and on 25 i got 1 but if you can't then thank you for helping!
ybarrap
  • ybarrap
Probabilities can not be greater than 1 so #27 is not 4
ybarrap
  • ybarrap
For #27 what do you have for P(B|A)=P(B and A)/P(A) ?
ybarrap
  • ybarrap
We already did #23
anonymous
  • anonymous
oh well you had .1/.4 so in 27 i just did .4/.1
ybarrap
  • ybarrap
No. The "|" does NOT mean division. It means "given" P(A|B) means Probability of event A given event B.
anonymous
  • anonymous
you did .1/.4
ybarrap
  • ybarrap
$$ P(B|A)=\cfrac{P(A\cap B)}{P(A)}\\ P(A|B)=\cfrac{P(A\cap B)}{P(B)} $$ Do you see the difference above between the two?
ybarrap
  • ybarrap
Yes and that one is correct
anonymous
  • anonymous
i guess I'm not understanding this then
ybarrap
  • ybarrap
So we are dealing with probabilities of events. When we are given that an event occurs, we are asked what is the probability that the other will occur. If they are independent, then there is no effect. But if they are not independent, then conditional probability tells us that Probability of A given B is the Probability of A and B divided by the probability of B. Here is a visual |dw:1433295585568:dw|
anonymous
  • anonymous
ok
ybarrap
  • ybarrap
That fraction is $$ P(A|B)=\cfrac{P(A\cap B)}{P(B)} $$
anonymous
  • anonymous
that looks like alien writing to me
ybarrap
  • ybarrap
$$ P(A|B)=P(A \text{ given }B)=\cfrac{P(A\cap B)}{P(B)}=\cfrac{P(A\text{ and }B)}{P(B)} $$ Is this better?
anonymous
  • anonymous
haha no I'm sorry... what does it look like with the values plugged in
ybarrap
  • ybarrap
|dw:1433296024303:dw|
ybarrap
  • ybarrap
$$ P(A|B)=P(A \text{ given }B)=\cfrac{P(A\cap B)}{P(B)}=\cfrac{P(A\text{ and }B)}{P(B)}=.1/.4 $$ For #26
ybarrap
  • ybarrap
For #27 $$ P(B|A)=P(B \text{ given }A)=\cfrac{P(A\cap B)}{P(A)}=\cfrac{P(A\text{ and }B)}{P(A)}=.1/.5 $$
anonymous
  • anonymous
ok so its .2
anonymous
  • anonymous
it just made a lot more sense
anonymous
  • anonymous
so because (a and b) the value is .1 and P(a)=.5
ybarrap
  • ybarrap
#24 $$ P(A|B)=P(A \text{ given }B)=\cfrac{P(A\cap B)}{P(B)}=\cfrac{P(A\text{ and }B)}{P(B)}=(0.3\times0.7)/0.7 $$ Notice that the result is independent of B because P(A|B) = P(A)
ybarrap
  • ybarrap
This happened because A and B are Independent
ybarrap
  • ybarrap
If you understood that, then you can do #25 without any calculations
ybarrap
  • ybarrap
Correct on your last comment
anonymous
  • anonymous
ok thank you i got it now
ybarrap
  • ybarrap
You're welcome

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