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S1 Probability help

Events A and B are independent with P(A) = 1/4 and P(AUB) = 2/3
Find:
b) P(B)
Independent so P(A)P(B) = P(ANB) therefore = 1/4P(B)
So P(A) + P(B) - P(ANB) = P(AUB)
Therefore 1/4 + 3/4P(B) = 2/3
Therefore 2/3 - 1/4 = 5/12
P(B) = 5/12 over 3/4 = 5/9 right??
c) P(A'|B) = ??
d) P(B'|A) = ??
Wait it's 5/12 and 1/9 respectively right??
Reply 2
Original post by AlphaArgonian
Events A and B are independent with P(A) = 1/4 and P(AUB) = 2/3
Find:
b) P(B)
Independent so P(A)P(B) = P(ANB) therefore = 1/4P(B)
So P(A) + P(B) - P(ANB) = P(AUB)
Therefore 1/4 + 3/4P(B) = 2/3
Therefore 2/3 - 1/4 = 5/12
P(B) = 5/12 over 3/4 = 5/9 right??


If you have 2/3 = 1/4 * P(B) - then how can you find P(B)?
Original post by Zacken
If you have 2/3 = 1/4 * P(B) - then how can you find P(B)?


Divide both sides by 1/4
Wait why?
(edited 8 years ago)
Reply 4
Original post by AlphaArgonian
Divide both sides by 1/4
Wait why?


And that's your answer, isn't it?
Reply 5
Oh, and: moved to maths.
Original post by Zacken
And that's your answer, isn't it?


Uhhh I get 8/3 from that
(edited 8 years ago)
Reply 7
Original post by AlphaArgonian
Divide both sides by 1/4
Wait why?


If ax=bax = b then: axa=ba    x=ba\displaystyle \frac{ax}{a} = \frac{b}{a} \iff x = \frac{b}{a}, in this case, a=14a=\frac{1}{4}.
Reply 8
Original post by AlphaArgonian
Uhhh I gett 8/3 from that


Argh!! Ignore me, I read your union and intersection signs wrong!!
Original post by Zacken
Argh!! Ignore me, I read your union and intersection signs wrong!!


Haha, that's fine I'll probably do that in the real exam this year
Reply 10
Original post by AlphaArgonian
Events A and B are independent with P(A) = 1/4 and P(AUB) = 2/3
Find:
b) P(B)
Independent so P(A)P(B) = P(ANB) therefore = 1/4P(B)
So P(A) + P(B) - P(ANB) = P(AUB)
Therefore 1/4 + 3/4P(B) = 2/3
Therefore 2/3 - 1/4 = 5/12
P(B) = 5/12 over 3/4 = 5/9 right??
c) P(A'|B) = ??
d) P(B'|A) = ??



Yeah, you get 5/9. That's correct.

For the next part, try looking at the conditional probability formula and recall how you can write P(A':wink: if you know P(A).
Original post by Zacken
Yeah, you get 5/9. That's correct.

For the next part, try looking at the conditional probability formula and recall how you can write P(A':wink: if you know P(A).


Ah thanks
1-P(A)?
Reply 12
Original post by AlphaArgonian
Ah thanks
1-P(A)?


Yep.
Original post by Zacken
Yep.


Cheers
Original post by Zacken
Yeah, you get 5/9. That's correct.

For the next part, try looking at the conditional probability formula and recall how you can write P(A':wink: if you know P(A).


Original post by AlphaArgonian
Cheers

.
Also, if A and B are independent, A' and B are also independent, so P(A'|B) = P(A')

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