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anonymous

  • one year ago

What is the marginal probability of X = 2?

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  1. anonymous
    • one year ago
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  2. hartnn
    • one year ago
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    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) = .. .

  3. hartnn
    • one year ago
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    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!

  4. anonymous
    • one year ago
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    Why 1 and 2 though? where did those values come from.. For the number 7, I just overcomplicated myself.

  5. anonymous
    • one year ago
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    and which formula do I have to plug them into? the same as before? 4/(9*x*y) = .

  6. hartnn
    • one year ago
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    Its given in Question 7, that they can take values 1 and 2. Q 7,8,9 are dependent

  7. hartnn
    • one year ago
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    yes, that formula applies for all those questions

  8. anonymous
    • one year ago
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    I didn't know they were dependent... They were confusing! `Usually they just repeat the question.

  9. hartnn
    • one year ago
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    True, it should be mentioned. just asking for marginal probability of X=2, is absurd!

  10. anonymous
    • one year ago
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    Yes that is! I was so confused. But I believe it's true

  11. hartnn
    • one year ago
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    so what did you get for question 7 and 8 ?

  12. anonymous
    • one year ago
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    For 7, I got 1/9 and the 8 , 1/3 Is that correct?

  13. hartnn
    • one year ago
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    absolutely :) need help with 9th one too?

  14. anonymous
    • one year ago
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    Yes please! Was just about to ask You're awsome by the way!

  15. hartnn
    • one year ago
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    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)

  16. hartnn
    • one year ago
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    so using the formula, we actually need, \(\Large \dfrac{P(Y=1 \quad and \quad X=1)}{P(X=1)}\) makes sense?

  17. hartnn
    • one year ago
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    the | is read as 'given'

  18. anonymous
    • one year ago
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    And P stands for Probability. Is it also refering again to the first Function from question 7?

  19. hartnn
    • one year ago
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    yes, and yes! the numerator is same as 7th question, with values changed. the denominator is same as 8th question, with values changed!

  20. anonymous
    • one year ago
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    So numeratior is 4/9

  21. hartnn
    • one year ago
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    correct :) get the denominator now

  22. anonymous
    • one year ago
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    2/3?

  23. hartnn
    • one year ago
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    yup, so whats the answer for 9th?

  24. anonymous
    • one year ago
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    1 :)

  25. hartnn
    • one year ago
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    the numerator is 4/9 and denominator is 2/3 try again?

  26. anonymous
    • one year ago
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    2/3 lol. Sorry

  27. hartnn
    • one year ago
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    \(\huge \checkmark\)

  28. anonymous
    • one year ago
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    Thanks a whole lot!!

  29. hartnn
    • one year ago
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    welcome ^_^ happy to help :)

  30. anonymous
    • one year ago
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    Can I ask one last question?

  31. anonymous
    • one year ago
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    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...

  32. hartnn
    • one year ago
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    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 \(\Large P(A \cap B) = P(A) \times P(B)\)

  33. hartnn
    • one year ago
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    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 :) )

  34. anonymous
    • one year ago
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    2/3 for P(x=1) ... so now finding y?

  35. anonymous
    • one year ago
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    Am I thinking wrong or is Y=1 the same? 2/3?

  36. hartnn
    • one year ago
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    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 :)

  37. anonymous
    • one year ago
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    so that's just 4/9 :) Easier than I thought. Should stop overcomplicating things!

  38. hartnn
    • one year ago
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    yup :) good luck!

  39. anonymous
    • one year ago
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    Thank you!!!

  40. hartnn
    • one year ago
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    \[ \begin{array}l\color{red}{\text{w}}\color{orange}{\text{e}}\color{#E6E600}{\text{l}}\color{green}{\text{c}}\color{blue}{\text{o}}\color{purple}{\text{m}}\color{purple}{\text{e}}\color{red}{\text{ }}\color{orange}{\text{^}}\color{#E6E600}{\text{_}}\color{green}{\text{^}}\color{blue}{\text{}}\end{array} \]

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