anonymous
  • anonymous
What is the marginal probability of X = 2?
Mathematics
jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
hartnn
  • hartnn
so, lets start with 7. X can take values: 1 ,2 Y can take values: 1 ,2 They asked what is f(x,y), when x =2 and y =2 so you just plug in x=2 and y =2 in f(x,y) 4/(9*2*2) = .. .
hartnn
  • hartnn
then we got to 8. They asked the probability when X takes the value =2. What about Y?? it can still take the value =1 OR =2 (It can't take both values at the same time!) so in f(x,y), plug in x=2, y=1 then plug in x=2,y=2 and just add them!

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anonymous
  • anonymous
Why 1 and 2 though? where did those values come from.. For the number 7, I just overcomplicated myself.
anonymous
  • anonymous
and which formula do I have to plug them into? the same as before? 4/(9*x*y) = .
hartnn
  • hartnn
Its given in Question 7, that they can take values 1 and 2. Q 7,8,9 are dependent
hartnn
  • hartnn
yes, that formula applies for all those questions
anonymous
  • anonymous
I didn't know they were dependent... They were confusing! `Usually they just repeat the question.
hartnn
  • hartnn
True, it should be mentioned. just asking for marginal probability of X=2, is absurd!
anonymous
  • anonymous
Yes that is! I was so confused. But I believe it's true
hartnn
  • hartnn
so what did you get for question 7 and 8 ?
anonymous
  • anonymous
For 7, I got 1/9 and the 8 , 1/3 Is that correct?
hartnn
  • hartnn
absolutely :) need help with 9th one too?
anonymous
  • anonymous
Yes please! Was just about to ask You're awsome by the way!
hartnn
  • hartnn
good learners like you make me awesome ^_^ So the definition of Conditional probability \(\Large P(A|B) = \dfrac{P(A\cap B)}{P(B)}\) we need, P(Y=1 | X=1)
hartnn
  • hartnn
so using the formula, we actually need, \(\Large \dfrac{P(Y=1 \quad and \quad X=1)}{P(X=1)}\) makes sense?
hartnn
  • hartnn
the | is read as 'given'
anonymous
  • anonymous
And P stands for Probability. Is it also refering again to the first Function from question 7?
hartnn
  • hartnn
yes, and yes! the numerator is same as 7th question, with values changed. the denominator is same as 8th question, with values changed!
anonymous
  • anonymous
So numeratior is 4/9
hartnn
  • hartnn
correct :) get the denominator now
anonymous
  • anonymous
2/3?
hartnn
  • hartnn
yup, so whats the answer for 9th?
anonymous
  • anonymous
1 :)
hartnn
  • hartnn
the numerator is 4/9 and denominator is 2/3 try again?
anonymous
  • anonymous
2/3 lol. Sorry
hartnn
  • hartnn
\(\huge \checkmark\)
anonymous
  • anonymous
Thanks a whole lot!!
hartnn
  • hartnn
welcome ^_^ happy to help :)
anonymous
  • anonymous
Can I ask one last question?
anonymous
  • anonymous
What would be the probability of X=1 and Y=1 assuming that X and Y are independent? -- with the same 4/9(xy) - Just don't understand what the independence has to do with anything...
hartnn
  • hartnn
Independent events means one even does not depend on other... so X being = 1, does not depend on whether Y=1 or 2 if 2 events A,B are independent, then \(\Large P(A \cap B) = P(A) \times P(B)\)
hartnn
  • hartnn
they are asking P(X=1 and Y=1) find P(X=1) and P(Y=1) separately, and just multiply them! (You already got P(X=1) in question 9 :) )
anonymous
  • anonymous
2/3 for P(x=1) ... so now finding y?
anonymous
  • anonymous
Am I thinking wrong or is Y=1 the same? 2/3?
hartnn
  • hartnn
it would be same, because if you interchange x and y, you still get same f(x,y)! so both P(X=1) = P(Y=1) = 2/3 just multiply them :)
anonymous
  • anonymous
so that's just 4/9 :) Easier than I thought. Should stop overcomplicating things!
hartnn
  • hartnn
yup :) good luck!
anonymous
  • anonymous
Thank you!!!
hartnn
  • hartnn
\[ \begin{array}l\color{red}{\text{w}}\color{orange}{\text{e}}\color{#E6E600}{\text{l}}\color{green}{\text{c}}\color{blue}{\text{o}}\color{purple}{\text{m}}\color{purple}{\text{e}}\color{red}{\text{ }}\color{orange}{\text{^}}\color{#E6E600}{\text{_}}\color{green}{\text{^}}\color{blue}{\text{}}\end{array} \]

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