Let $f(x,y ) =\dfrac{x+y}{32}$, x =1,2; y = 1,2,3,4
find $P (1\leq Y\leq 3| X =1)$
Please, help

Question

Let $f(x,y ) =\dfrac{x+y}{32}$,    x =1,2;   y = 1,2,3,4
find $P (1\leq Y\leq 3| X =1)$ 
Please, help

loser66 · Answer

I have P (X =1) = 14/32

loser66 · Answer

need help on the first part, how to interpret it?

loser66 · Answer

Don't we have to apply Baye's theorem here?

loser66 · Answer

I mean:$P(Y=1|X=1) =\dfrac{P(Y=1,X=1)}{P(X=1)}$

tkhunny · Answer

Yes.

loser66 · Answer

If it is so, then, the first one is (2/32 )/(14/32)= 1/7 different from your answer

loser66 · Answer

So, we just add them up?

tkhunny · Answer

Yup.  I get P(Y=1|X=1) = 1/7

loser66 · Answer

Yes, I got it, thank you so much :)

tkhunny · Answer

Very confusing typos.  I'll delete that.

loser66 · Answer

One more question: how can you get it quickly like that? I have to calculate the probability and fill up the table, calculate everything step by step while you got it right after you see it. Is there any tips?

tkhunny · Answer

Not really.  Use all the tools available to you.

loser66 · Answer

OH,,, if it is so, thanks but I can't. On test, we are not allowed to use technology. :)

tkhunny · Answer

Well, spread the table and fill it in!  Did you also check to see if it actually qualified as a Probability Distribution?

loser66 · Answer

What do you mean ?

tkhunny · Answer

Does $\sum{f(Everything)} = 1$?