Urn I and Urn II each have 2 red chips and 2 white chips. Two chips are drawn simultaneously from 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
Pdf of X1 = pdf of X2
=(1/2,1/2) 0≤X1≤1, X1∈Z
= 0 otherwise
Pdf of X = X1 and X2
= (1/4, 1/2, 1/4) 0≤X≤2, X∈Z
= 0 otherwise
Please explain?
To find the pdf (probability density function) of X1 + X2, we need to consider all the possible outcomes and calculate their probabilities.
First, let's determine all the possible values of X1 + X2:
- If both chips drawn from both urns are white, X1 + X2 = 0 (no red chips).
- If one chip drawn from each urn is white and the other is red, X1 + X2 = 1 (one red chip).
- If both chips drawn from both urns are red, X1 + X2 = 2 (two red chips).
Next, let's calculate the probabilities for each case:
Case 1: X1 + X2 = 0
The probability of drawing two white chips from both urns is:
P(X1 = 0) * P(X2 = 0) = (2/4 * 1/3) * (2/4 * 1/3) = 1/18
Case 2: X1 + X2 = 1
The probability of drawing one red chip from Urn I and one white chip from Urn II is:
P(X1 = 1) * P(X2 = 0) = (2/4 * 1/3) * (2/4 * 2/3) = 4/36
The probability of drawing one white chip from Urn I and one red chip from Urn II is the same:
P(X1 = 0) * P(X2 = 1) = (2/4 * 2/3) * (2/4 * 1/3) = 4/36
Therefore, the total probability for X1 + X2 = 1 is 4/36 + 4/36 = 8/36 = 2/9
Case 3: X1 + X2 = 2
The probability of drawing two red chips from both urns is:
P(X1 = 2) * P(X2 = 2) = (2/4 * 2/3) * (2/4 * 1/3) = 4/18
Adding up the probabilities for all three cases, we have:
P(X1 + X2 = 0) = 1/18
P(X1 + X2 = 1) = 2/9
P(X1 + X2 = 2) = 4/18 = 2/9
So, the pdf of X1 + X2 is:
P(X1 + X2 = 0) = 1/18
P(X1 + X2 = 1) = 2/9
P(X1 + X2 = 2) = 2/9
To find the probability density function (pdf) of X1 + X2, we need to consider all possible outcomes of X1 and X2 and calculate their probabilities.
First, let's list all possible outcomes of X1 and X2:
X1 = 0, X2 = 0: Both samples have 0 red chips.
X1 = 0, X2 = 1: The first sample has 0 red chips, and the second sample has 1 red chip.
X1 = 1, X2 = 0: The first sample has 1 red chip, and the second sample has 0 red chips.
X1 = 1, X2 = 1: Both samples have 1 red chip.
X1 = 2, X2 = 0: The first sample has 2 red chips, and the second sample has 0 red chips.
X1 = 0, X2 = 2: The first sample has 0 red chips, and the second sample has 2 red chips.
X1 = 2, X2 = 1: The first sample has 2 red chips, and the second sample has 1 red chip.
X1 = 1, X2 = 2: The first sample has 1 red chip, and the second sample has 2 red chips.
X1 = 2, X2 = 2: Both samples have 2 red chips.
Next, let's find the probability of each outcome:
P(X1 = 0, X2 = 0) = P(no red chips in the first sample) * P(no red chips in the second sample)
= (C(2, 0) * C(2, 2)) / C(4, 2) * (C(2, 0) * C(2, 2)) / C(4, 2)
= (1 * 1) / 6 * (1 * 1) / 6
= 1/36
Similarly, we can calculate the probabilities for each outcome:
P(X1 = 0, X2 = 1) = 2/36
P(X1 = 1, X2 = 0) = 2/36
P(X1 = 1, X2 = 1) = 4/36
P(X1 = 2, X2 = 0) = 1/36
P(X1 = 0, X2 = 2) = 1/36
P(X1 = 2, X2 = 1) = 2/36
P(X1 = 1, X2 = 2) = 2/36
P(X1 = 2, X2 = 2) = 1/36
Finally, we can write the pdf of X1 + X2 using the calculated probabilities:
pdf(X1 + X2 = 0) = P(X1 = 0, X2 = 0) = 1/36
pdf(X1 + X2 = 1) = P(X1 = 0, X2 = 1) + P(X1 = 1, X2 = 0) = 2/36 + 2/36 = 4/36
pdf(X1 + X2 = 2) = P(X1 = 1, X2 = 1) + P(X1 = 2, X2 = 0) + P(X1 = 0, X2 = 2) = 4/36 + 1/36 + 1/36 = 6/36
pdf(X1 + X2 = 3) = P(X1 = 2, X2 = 1) + P(X1 = 1, X2 = 2) = 2/36 + 2/36 = 4/36
pdf(X1 + X2 = 4) = P(X1 = 2, X2 = 2) = 1/36
So the pdf of X1 + X2 is:
pdf(X1 + X2 = 0) = 1/36
pdf(X1 + X2 = 1) = 4/36
pdf(X1 + X2 = 2) = 6/36
pdf(X1 + X2 = 3) = 4/36
pdf(X1 + X2 = 4) = 1/36