Thread: joint density function

An unbiased coin is tossed three times. Let X be the number of heads on the first toss and Y the number of heads on the second and third tosses combined. Find the joint density function of X and Y.

Hmm i reckon I'm wrong but is the correct answer 1/2?

SOrry i don't know how to draw a table here but all i did for this question was add up the row total??
so 1/8 + 1/4 + 1/8

_____Y______
_____0____1_____2
0 |....1/8...1/4....1/8
1 |....1/8...1/4...1/8 | X


P(X = 0, Y=0) = 1/8
P(X=0, Y=1) = 1/4
P(X=0,Y=2) = 1/8

which equals to 1/2... is this what I'm supposed to do or am i way off(highly likely)?

Yeahhh and here are the possibilities i drew up:
____X Y
HHH 1 2
HHT 1 1
HTH 1 1
THH 0 2
HTT 1 0
THT 0 1
TTH 0 1
TTT 0 0

Originally Posted by gconfused

An unbiased coin is tossed three times. Let X be the number of heads on the first toss and Y the number of heads on the second and third tosses combined. Find the joint density function of X and Y.

Hmm i reckon I'm wrong but is the correct answer 1/2? Mr F says: Answer to what?? The joint density function is the table you drew below (which is correct, by the way). I don't see any other question in your post ....?

SOrry i don't know how to draw a table here but all i did for this question was add up the row total??
so 1/8 + 1/4 + 1/8

_____Y______
_____0____1_____2
0 |....1/8...1/4....1/8
1 |....1/8...1/4...1/8 | X


P(X = 0, Y=0) = 1/8
P(X=0, Y=1) = 1/4
P(X=0,Y=2) = 1/8

which equals to 1/2... is this what I'm supposed to do or am i way off(highly likely)?

Yeahhh and here are the possibilities i drew up:
____X Y
HHH 1 2
HHT 1 1
HTH 1 1
THH 0 2
HTT 1 0
THT 0 1
TTH 0 1
TTT 0 0

..

oh im so dumb i thought i was supposed to get a value haha