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Original post by justinawe
Hmm?

You understand that m at Q is 12-\dfrac{1}{2} ?

You understand this means that dydx\dfrac{dy}{dx} at Q is 12-\dfrac{1}{2} ?

You understand that y=14x2dydx=12xy = \dfrac{1}{4}x^2 \Rightarrow \dfrac{dy}{dx} = \dfrac{1}{2}x ?

and thus 12x=12\dfrac{1}{2}x = -\dfrac{1}{2} ?


Ohhhhh..... OK... !

its just a revers - ie thing

I got it! /// Wow!! - Your maths skills are comparable to a DON!!

i would rep but im out :colondollar:
thank you!

ryan
Reply 1741
Original post by purplemind
It would be really nice. :tongue:


20130413_155934.jpg 20130413_160108.jpg
Original post by tigerz
20130413_155934.jpg 20130413_160108.jpg

Thank you.
The attached document looks quite useful. :biggrin:
Reply 1743
Original post by purplemind
Thank you.
The attached document looks quite useful. :biggrin:


Haha no problem :smile:
Original post by tigerz
20130413_155934.jpg 20130413_160108.jpg


Daymnn guuurl loving those colorful notes :wink:
Original post by ryanb97
Ohhhhh..... OK... !

its just a revers - ie thing

I got it! /// Wow!! - Your maths skills are comparable to a DON!!

i would rep but im out :colondollar:
thank you!

ryan


Oh come now, I'm just more experienced :tongue: you're in year 10 or 11 iirc...
Original post by tigerz
Haha no problem :smile:

Anyway, I'm being a bit thick now, but seriosly, I don't know how to go about doing it.
I am given P(A)=3/4
P(B|A)=1/5
P(B'|A')=4/7
They want me to find P(A and B)- that's easy, I did it. But then they want P(B). And it confuses me a bit.
DSC_0225.jpg
Reply 1747
c2 friday aaaaarghhh
Original post by Boy_wonder_95
Daymnn guuurl loving those colorful notes :wink:


I didn't realise you were american :hmmm:
Original post by purplemind

They want me to find P(A and B)- that's easy, I did it. But then they want P(B). And it confuses me a bit.


Surely the P(B)=P(BA)+P(BA)\displaystyle P(B) = P(B|A) + P(B|A') ?

EDIT: I meant P(B)=P(AB)+P(AB)\displaystyle P(B) = P(A \cap B) + P(A' \cap B)
(edited 10 years ago)
Original post by Scientific Eye
Surely the P(B)=P(BA)+P(BA)\displaystyle P(B) = P(B|A) + P(B|A') ?

But why?
Even if they are not mutally exclusive?
(edited 10 years ago)
Original post by justinawe
Oh come now, I'm just more experienced :tongue: you're in year 10 or 11 iirc...


year 11 .... but i must say i havent been in a proper maths lesson in 4 years :cry2:

and i wont be next year...:cry:

lol

ryan
Original post by justinawe
I didn't realise you were american :hmmm:


Haha, is that a bad thing? :hmmm:
Reply 1753
Original post by Boy_wonder_95
Daymnn guuurl loving those colorful notes :wink:

Original post by justinawe
I didn't realise you were american :hmmm:


LOOL! IKR thank youuu, but I only just realised they have loads of spelling errors :colondollar: btw, we're literally meeting on every thread :laugh:
(edited 10 years ago)
Original post by Boy_wonder_95
Haha, is that a bad thing? :hmmm:


No, of course not. Are you really, though? :tongue:
Original post by justinawe
No, of course not. Are you really, though? :tongue:


Nah :tongue:, it's just that TSR has 'colourful' underlined in red which buggers me :rolleyes:
Original post by purplemind
But why?


We are only concerned with the outcomes where event B\displaystyle B occurs.

Look at the probability tree. You have event A\displaystyle A and event B\displaystyle B. We don't care if event A\displaystyle A happens or not, only that event B\displaystyle B does happen. Hence we find all the ends of the probability tree where event B\displaystyle B is an outcome and then add them up.

That is, if I'm not incorrect in my thinking.
Original post by purplemind
Anyway, I'm being a bit thick now, but seriosly, I don't know how to go about doing it.
I am given P(A)=3/4
P(B|A)=1/5
P(B'|A')=4/7
They want me to find P(A and B)- that's easy, I did it. But then they want P(B). And it confuses me a bit.
DSC_0225.jpg


Use these two formulae (which are in the formulae booklet):


P(AB)=P(A)P(BA)P(A \cap B) = P(A)P(B | A)

P(AB)=P(A)+P(B)P(AB)P(A \cup B) = P(A) + P(B) - P(A \cap B)


Find P(AB)P(A \cap B) using the first formula, then sub it into the second one.
(edited 10 years ago)
Original post by tigerz
LOOL! IKR, but I only just realised they have loads of spelling errors :colondollar: btw, we're literally meeting on every thread :laugh:


Still they make my S1 notes look rusty in comparison :redface:, who me or justin?
Original post by Scientific Eye
We are only concerned with the outcomes where event B\displaystyle B occurs.

Look at the probability tree. You have event A\displaystyle A and event B\displaystyle B. We don't care if event A\displaystyle A happens or not, only that event B\displaystyle B does happen. Hence we find all the ends of the probability tree where event B\displaystyle B is an outcome and then add them up.

That is, if I'm not incorrect in my thinking.


Ok, so P(B|A)=1/5
P(B|A')=3/7
I add them and get P(B)=22/35
My textbook says it should be 9/35
So I don't know if it is right and the answer in the textbook is wrong or the other way round. :confused:

Original post by justinawe
Use these to formulae (which are in the formulae booklet):


P(AB)=P(A)P(BA)P(A \cap B) = P(A)P(B | A)

P(AB)=P(A)+P(B)P(AB)P(A \cup B) = P(A) + P(B) - P(A \cap B)


Find P(AB)P(A \cap B) using the first formula, then sub it into the second one.

I still don't know how to find P(A U B) from what I am given. :<
(edited 10 years ago)

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