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anonymous
 one year ago
What is the marginal probability of X = 2?
anonymous
 one year ago
What is the marginal probability of X = 2?

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hartnn
 one year ago
Best ResponseYou've already chosen the best response.5so, 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
 one year ago
Best ResponseYou've already chosen the best response.5then 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!

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Why 1 and 2 though? where did those values come from.. For the number 7, I just overcomplicated myself.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0and which formula do I have to plug them into? the same as before? 4/(9*x*y) = .

hartnn
 one year ago
Best ResponseYou've already chosen the best response.5Its given in Question 7, that they can take values 1 and 2. Q 7,8,9 are dependent

hartnn
 one year ago
Best ResponseYou've already chosen the best response.5yes, that formula applies for all those questions

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I didn't know they were dependent... They were confusing! `Usually they just repeat the question.

hartnn
 one year ago
Best ResponseYou've already chosen the best response.5True, it should be mentioned. just asking for marginal probability of X=2, is absurd!

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Yes that is! I was so confused. But I believe it's true

hartnn
 one year ago
Best ResponseYou've already chosen the best response.5so what did you get for question 7 and 8 ?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0For 7, I got 1/9 and the 8 , 1/3 Is that correct?

hartnn
 one year ago
Best ResponseYou've already chosen the best response.5absolutely :) need help with 9th one too?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Yes please! Was just about to ask You're awsome by the way!

hartnn
 one year ago
Best ResponseYou've already chosen the best response.5good learners like you make me awesome ^_^ So the definition of Conditional probability \(\Large P(AB) = \dfrac{P(A\cap B)}{P(B)}\) we need, P(Y=1  X=1)

hartnn
 one year ago
Best ResponseYou've already chosen the best response.5so using the formula, we actually need, \(\Large \dfrac{P(Y=1 \quad and \quad X=1)}{P(X=1)}\) makes sense?

hartnn
 one year ago
Best ResponseYou've already chosen the best response.5the  is read as 'given'

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0And P stands for Probability. Is it also refering again to the first Function from question 7?

hartnn
 one year ago
Best ResponseYou've already chosen the best response.5yes, and yes! the numerator is same as 7th question, with values changed. the denominator is same as 8th question, with values changed!

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0So numeratior is 4/9

hartnn
 one year ago
Best ResponseYou've already chosen the best response.5correct :) get the denominator now

hartnn
 one year ago
Best ResponseYou've already chosen the best response.5yup, so whats the answer for 9th?

hartnn
 one year ago
Best ResponseYou've already chosen the best response.5the numerator is 4/9 and denominator is 2/3 try again?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Thanks a whole lot!!

hartnn
 one year ago
Best ResponseYou've already chosen the best response.5welcome ^_^ happy to help :)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Can I ask one last question?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0What 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
 one year ago
Best ResponseYou've already chosen the best response.5Independent 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
 one year ago
Best ResponseYou've already chosen the best response.5they 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
 one year ago
Best ResponseYou've already chosen the best response.02/3 for P(x=1) ... so now finding y?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Am I thinking wrong or is Y=1 the same? 2/3?

hartnn
 one year ago
Best ResponseYou've already chosen the best response.5it 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
 one year ago
Best ResponseYou've already chosen the best response.0so that's just 4/9 :) Easier than I thought. Should stop overcomplicating things!

hartnn
 one year ago
Best ResponseYou've already chosen the best response.5\[ \begin{array}l\color{red}{\text{w}}\color{orange}{\text{e}}\color{#E6E600}{\text{l}}\color{green}{\text{c}}\color{blue}{\text{o}}\color{purple}{\text{m}}\color{purple}{\text{e}}\color{red}{\text{ }}\color{orange}{\text{^}}\color{#E6E600}{\text{_}}\color{green}{\text{^}}\color{blue}{\text{}}\end{array} \]
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