pdf of chips drawn from urn 1 and urn 2 Urn I and urn II each have two red chips and two white chips. Two chips are drawn simultaneously from each urn. Let X1 be he number of red chips in the first sample and X2 be the number of red chips in the second sample. Find the pdf of X1 + X2.
solution: 0: 1/36, 1: 2/9, 2: 1/2, 3: 2/9, 4: 1/36.
can someone help me understand this problem
 A: The draw from each urn is two red with probability $\frac 16$, because to get two red you need to star with a red (probability $\frac 12$) and then get the other red (probability $\frac 13)$.  Similarly, you get two white with probability $\frac 16$, so you get one of each with probability $\frac 23$.  Now to get four reds you need to get two reds out of each urn.  If you want two of each you have a number of ways to get there-add them up.
A: $R1$ $R2$ $W1$ $ W2 $ --Urn I
$R3$ $ R4 $ $W3 $ $ W4$ --Urn II
you draw $4$ chips - $2$ from first urn and $2$ from second..How many combinations are there?
$C(4,2)*C(4,2)=36$ different combinations--
Well , among them only one have $4$ reds.
Only one have $4$ whites or $0$ reds.
1 red and 3 white will be 8 cases:
$R1$ $ W1$ $  W3 $ $W4 $   ||$  R2 ...          ||  R3...  $   ||   R4...
$R1$ $ W2$ $  W3 $ $W4 $  ||  $  R2...              ||  R3...   $  ||    R4...
By symmetry,  3 red and 1 white will be 8 cases too..
And the remaining 18 combinations will be 2 red and 2 white.
Hence , we get your pdf of red chips in the total sample.
