Hard for us to say when you don't show us what you came up with for the joint probability distribution.
Hi again! Have a bit of a problem with this question and would gratefully appreciate some help.
There are 4 cards in a game, numbered 1 to 4. A player randomly draws one card at a time (and without replacement). The game ends when the numbers on the drawn cards sum to 3 and above.
Let X be a random variable - The sum of numbers on the drawn cards.
Let Y be a RV - The number of cards the player had drawn.
Find the joint probability of X & Y.
Well, my struggle is mostly in which method do I calculate the probabilities. I know that X can take the values 3,4,5,6 and Y can take 1,2.
The sample space is 8 I think, because there are 8 possible scenarios of taking cards out until we stop. Am I right to think this way?
However when putting the probabilities in the table, they sum up to 6/8 only. What am I overlooking?
Thanks a bunch!
Hard for us to say when you don't show us what you came up with for the joint probability distribution.
I don't have emotions and sometimes that makes me very sad.
It is a good start to know that the support of the joint distribution is a subset of the product set of individual supports.
So in your case there are at most support points. As what Dason suggested, show us the joint pmf calculated by you on those points (list it or tabulate it) and we can check which part goes wrong.
My apologies. Here goes-
P(x=3,y=1)= 1/8 (because that would mean choosing the 3 card so there's only one way to do that)
P(x=3,y=2)=2/8 (either taking card '1' out and then card '2' or the other way around)
P(x=4, y=2)=2/8
P(x=5,y=1);P(x=6,y=1)=0
P(x=5,y=2)=2/8
P(x=6,y=2)=1/8
And I just realized I found my mistake. Thank you all!
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