Urn I and urn II each have two red chips and two white chips. Two chips are drawn simultaneously form each urn. Let X1 be the number of red chips in the first sample and X2 the number of red chips in the second sample. Find the pdf of X1+X2.
To find the probability distribution function (pdf) of X1+X2, let's consider all possible outcomes of drawing chips from urn I and urn II.
From urn I, the possible outcomes are:
1. Drawing 2 red chips: P(2 red chips) = (2/4) * (1/3) = 1/6
2. Drawing 1 red chip and 1 white chip: P(1 red, 1 white) = (2/4) * (2/3) = 4/12 = 1/3
3. Drawing 2 white chips: P(2 white chips) = (2/4) * (1/3) = 1/6
From urn II, the possible outcomes are the same as from urn I.
Now, let's calculate the probability distribution of X1+X2:
When X1+X2=0, it means both samples have 0 red chips. The probability is the product of the probabilities of getting 2 white chips from both urns:
P(X1+X2=0) = (1/6) * (1/6) = 1/36
When X1+X2=1, it means one sample has 1 red chip and the other has 0 red chips. The probability is the sum of the probabilities of all possible combinations:
P(X1+X2=1) = (1/6) * (4/6) + (1/6) * (4/6) = 4/36 = 1/9
When X1+X2=2, it means both samples have 1 red chip. The probability is the product of the probabilities of getting 1 red chip from both urns:
P(X1+X2=2) = (1/6) * (1/6) = 1/36
When X1+X2=3, it means one sample has 2 red chips and the other has 1 red chip. The probability is the sum of the probabilities of all possible combinations:
P(X1+X2=3) = (1/6) * (1/6) + (4/36) + (1/6) * (1/6) = 7/36
When X1+X2=4, it means both samples have 2 red chips. The probability is the product of the probabilities of getting 2 red chips from both urns:
P(X1+X2=4) = (1/6) * (1/6) = 1/36
Therefore, the pdf of X1+X2 is:
P(X1+X2=0) = 1/36
P(X1+X2=1) = 1/9
P(X1+X2=2) = 1/36
P(X1+X2=3) = 7/36
P(X1+X2=4) = 1/36