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June 20, 2019, 03:20:40 pm AuthorTopic: Help needed...probability  (Read 317 times) Tweet Share

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Perla

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« on: October 08, 2018, 08:22:55 pm »
0
Hi everyone,

I don't know what happen, less than one month before the exam and I feel completely lost. I could do things really easily and now I cannot, maybe the stress or I don't know.. Do you h ave any advice or something, I do not want to fail my exam.

Also, if someone could help me with these 2 questions please.

If Pr (E ∩ F) = 0.4 , Pr (E∩ F) = 0.05 and (E ∩ F) is denoted as x, given that Pr (F | E) = 1/5
i. Find Pr (E ∪ F)
ii. Find Pr (E | F)

Thank you

LifeisaConstantStruggle

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• Respect: +82 Re: Help needed...probability
« Reply #1 on: October 08, 2018, 08:41:24 pm »
+1
It's best if you draw a Venn diagram for it, with E as a circle intersecting F, and a background.

Pr(E ∩ F) would be E minus the bit that intersects with F, so that would be 0.4
Pr(E' ∩ F) would be the background itself, and that is 0.05
Pr (F|E) = Pr(F∩E)/Pr(E) = x/Pr(E) = 1/5 so Pr(E) = 5x. Pr(E) is also 0.4+x if you look at the Venn diagram, so 5x = 0.4+x, x =Pr(F∩E) = 0.1
Pr(F) = 1-Pr(E ∩ F)-Pr(E' ∩ F) = 0.55 if you look at your Venn diagram again.
So that leaves us with the questions
Pr(E∪F)=Pr(E)+Pr(F)-Pr(F∩E)=0.5+0.55-0.1=0.95
Pr(E|F)=Pr(F∩E)/Pr(F')=0.1/0.45=0.222..
2016-2017: VCE (ATAR: 99.3)
2018: Monash.

Brittank88

• • Posts: 16
• Respect: 0 Re: Help needed...probability
« Reply #2 on: October 11, 2018, 06:25:02 pm »
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It's best if you draw a Venn diagram for it, with E as a circle intersecting F, and a background.

Pr(E ∩ F) would be E minus the bit that intersects with F, so that would be 0.4
Pr(E' ∩ F) would be the background itself, and that is 0.05
Pr (F|E) = Pr(F∩E)/Pr(E) = x/Pr(E) = 1/5 so Pr(E) = 5x. Pr(E) is also 0.4+x if you look at the Venn diagram, so 5x = 0.4+x, x =Pr(F∩E) = 0.1
Pr(F) = 1-Pr(E ∩ F)-Pr(E' ∩ F) = 0.55 if you look at your Venn diagram again.
So that leaves us with the questions
Pr(E∪F)=Pr(E)+Pr(F)-Pr(F∩E)=0.5+0.55-0.1=0.95
Pr(E|F)=Pr(F∩E)/Pr(F')=0.1/0.45=0.222..

Not to doubt you (more I want to see if I've made an error myself), but using a Karnaugh table I actually get slightly different results (Pr(F'∩E) for me is 0.4 not 0.1). I'm also confused as to why you compute Pr (F|E) when it's not one of OP's listed questions, and define Pr(E|F) as Pr(F∩E)/Pr(F') when I would have thought it should be Pr(F'∩E)/Pr(F'). The other thing that confuses me is that to answer both i and ii, you don't even have to touch the unknown value. Pr(E ∪ F) (which becomes 1-Pr(F'∩E')) has a numerical probability from the very beginning of 1-0.05 = 0.95, and so does Pr (E | F)=Pr(F'∩E)/Pr(F'), with a probability of 0.4/0.45=40/45=8/9.

Ultimately, more than happy to be proven wrong, getting the right answer is more important than being 'right'  I\'m Not A Robot

• Trailblazer
• • Posts: 26
• Respect: 0 Re: Help needed...probability
« Reply #3 on: October 11, 2018, 11:29:51 pm »
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Not to doubt you (more I want to see if I've made an error myself), but using a Karnaugh table I actually get slightly different results (Pr(F'∩E) for me is 0.4 not 0.1). I'm also confused as to why you compute Pr (F|E) when it's not one of OP's listed questions, and define Pr(E|F) as Pr(F∩E)/Pr(F') when I would have thought it should be Pr(F'∩E)/Pr(F'). The other thing that confuses me is that to answer both i and ii, you don't even have to touch the unknown value. Pr(E ∪ F) (which becomes 1-Pr(F'∩E')) has a numerical probability from the very beginning of 1-0.05 = 0.95, and so does Pr (E | F)=Pr(F'∩E)/Pr(F'), with a probability of 0.4/0.45=40/45=8/9.

Ultimately, more than happy to be proven wrong, getting the right answer is more important than being 'right' (Image removed from quote.)
I think the values you got were correct. It should be 0.4/0.45 and not 0.1/0.45.

Computing Pr (F|E)= 0.2 helps to work out the probability of all of 'E' which ends up being 5x when worked out algebraically.