4. (25 points) A gambler is playing a card game. On her turn, she drew three car

Question

4. (25 points) A gambler is playing a card game. On her turn, she drew three cards from the 12 remaining cards on the table. Because she has been counting the cards, she knew that there are 5 diamonds, 3 clubs, 2 hearts and 2 spades remaining on the table. Let X be the number of diamonds and Y be the number of hearts among three drawn cards. (a) Find the joint distribution of X and Y, f(x, y). Clearly specify the values of (x,y) that have nonzero probabilities (b) Find the marginal distribution of Y (compute f(y) for each nonzero y value) (c) Find the cumulative distribution function of Y, F().

Explanation / Answer

Here Y can take values 0, 1,2 (cecuase there is only 2 hearts) and X can take values 0, 1,2 , 3.

Number of ways of selecting 3 cards ou tof 12 is C(12, 3) = 220.

Out of 12 cards, 5 are diamonds, 2 are heart and 5 are other.

(a)

Number of ways of selecting 0 heart and 0 diamond is C(5,0)C(2,0)C(5,3) = 10

So

P(X=0, Y=0) = 10 / 220

Like wise

P(X=0,Y=1) = [C(5,0)C(2,1)C(5,2) ] /220 = 20/220

P(X=0,Y=2) = [C(5,0)C(2,2)C(5,1) ] /220 = 5/220

P(X=1,Y=0) = [C(5,1)C(2,0)C(5,2) ] /220 = 50/220

P(X=1,Y=1) = [C(5,1)C(2,1)C(5,1) ] /220 = 50/220

P(X=1,Y=2) = [C(5,1)C(2,2)C(5,0) ] /220 = 5/220

P(X=2,Y=0) = [C(5,2)C(2,0)C(5,1) ] /220 = 50/220

P(X=2,Y=1) = [C(5,2)C(2,1)C(5,0) ] /220 = 20/220

P(X=3,Y=0) =[C(5,3)C(2,0)C(5,0) ] /220 = 10/220

Rest all probabilities will be zero.

Following is the joint pdf in table format:

(b)

Following table shows the marginal pdf of Y:

(c)

The cumulative distribution function F(y) is

X 0 1 2 3 P(Y=y) 0 1/22 5/22 5/22 1/22 12/22 Y 1 2/22 5/22 2/22 0 9/22 2 5/220 5/220 0 0 1/22 P(X=x) 35/220 105/220 7/22 1/22 1

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