What is the marginal probability of X = 2?

- anonymous

What is the marginal probability of X = 2?

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- anonymous

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- 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

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

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

- anonymous

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

- hartnn

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

- hartnn

yes, that formula applies for all those questions

- anonymous

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

- hartnn

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

- anonymous

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

- hartnn

so what did you get for question 7 and 8 ?

- anonymous

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

- hartnn

absolutely :)
need help with 9th one too?

- anonymous

Yes please! Was just about to ask
You're awsome by the way!

- 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

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

- hartnn

the |
is read as 'given'

- anonymous

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

- 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

So numeratior is 4/9

- hartnn

correct :)
get the denominator now

- anonymous

2/3?

- hartnn

yup,
so whats the answer for 9th?

- anonymous

1 :)

- hartnn

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

- anonymous

2/3 lol. Sorry

- hartnn

\(\huge \checkmark\)

- anonymous

Thanks a whole lot!!

- hartnn

welcome ^_^
happy to help :)

- anonymous

Can I ask one last question?

- 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

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

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

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

- anonymous

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

- 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

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

- hartnn

yup :)
good luck!

- anonymous

Thank you!!!

- hartnn

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