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

1. anonymous

2. 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) = .. .

3. 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!

4. anonymous

Why 1 and 2 though? where did those values come from.. For the number 7, I just overcomplicated myself.

5. anonymous

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

6. hartnn

Its given in Question 7, that they can take values 1 and 2. Q 7,8,9 are dependent

7. hartnn

yes, that formula applies for all those questions

8. anonymous

I didn't know they were dependent... They were confusing! `Usually they just repeat the question.

9. hartnn

True, it should be mentioned. just asking for marginal probability of X=2, is absurd!

10. anonymous

Yes that is! I was so confused. But I believe it's true

11. hartnn

so what did you get for question 7 and 8 ?

12. anonymous

For 7, I got 1/9 and the 8 , 1/3 Is that correct?

13. hartnn

absolutely :) need help with 9th one too?

14. anonymous

15. 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)

16. hartnn

so using the formula, we actually need, $$\Large \dfrac{P(Y=1 \quad and \quad X=1)}{P(X=1)}$$ makes sense?

17. hartnn

the | is read as 'given'

18. anonymous

And P stands for Probability. Is it also refering again to the first Function from question 7?

19. hartnn

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

So numeratior is 4/9

21. hartnn

correct :) get the denominator now

22. anonymous

2/3?

23. hartnn

yup, so whats the answer for 9th?

24. anonymous

1 :)

25. hartnn

the numerator is 4/9 and denominator is 2/3 try again?

26. anonymous

2/3 lol. Sorry

27. hartnn

$$\huge \checkmark$$

28. anonymous

Thanks a whole lot!!

29. hartnn

welcome ^_^ happy to help :)

30. anonymous

Can I ask one last question?

31. 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...

32. 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 $$\Large P(A \cap B) = P(A) \times P(B)$$

33. 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 :) )

34. anonymous

2/3 for P(x=1) ... so now finding y?

35. anonymous

Am I thinking wrong or is Y=1 the same? 2/3?

36. 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 :)

37. anonymous

so that's just 4/9 :) Easier than I thought. Should stop overcomplicating things!

38. hartnn

yup :) good luck!

39. anonymous

Thank you!!!

40. hartnn

$\begin{array}l\color{red}{\text{w}}\color{orange}{\text{e}}\color{#E6E600}{\text{l}}\color{green}{\text{c}}\color{blue}{\text{o}}\color{purple}{\text{m}}\color{purple}{\text{e}}\color{red}{\text{ }}\color{orange}{\text{^}}\color{#E6E600}{\text{_}}\color{green}{\text{^}}\color{blue}{\text{}}\end{array}$