Question

Assume that you are asked to select three cards without replacement from the 39 cards that contain the hearts, diamonds, and clubs from an ordinary deck of 52 playing cards. Let X be the number of clubs selected and Y the number of diamonds. (a) Find the joint probability distribution of X and Y. (b) Find P[(X,Y)EA), where A is the region given by {(x,y) | X + y2 2} (a) Complete the joint probability distribution below. (Type integers or simplified fractions.) X f(x,y) 0 2 3 y 0 1 2 JU 3 (b) P[(X,Y)=A] =(Type an integer or simplified fraction.)

Answer 1

solution:

Given data

Total no.of cards = 39

No.of hearts = No.of Diamonds = No.of clubs = 13

when we take 3 cards,we have

n(S) = 39C3 = 9139

a) Let X be the no.of clubs selected and Y be the no.of diamonds selected

Then possible values of X nd Y are : X = { 0,1,2,3} , Y = { 0,1,2,3}

To find probability distribution

n(0,0) = 13C0 * 13C0 * 13C3 = 286

    n(0,1) = 13C0 * 13C1 * 13C2 = 1014

    n(0,2) = 13C0 * 13C2 * 13C1 = 1014

    n(0,3) = 13C0 * 13C3   = 286

    n(1,0) = 13C1 * 13C0 * 13C2 = 1014

   n(1,1) = 13C1 * 13C1 * 13C1 = 2197

    n(1,2) = 13C1 * 13C2   = 1014

   n(1,3) = 0 [ since we take only 3 cards ]

   n(2,0) = 13C2 * 13C0 * 13C1 = 1014

    n(2,1) = 13C2 * 13C1   = 1014

    n(2,2) = 0 [ since we take only 3 cards ]

    n(2,3) = 0    [ since we take only 3 cards ]

   n(3,0) = 13C3 * 13C0 = 286

   n(3,1) = 0 [ since we take only 3 cards ]

   n(3,2) = 0 [ since we take only 3 cards ]

   n(3,3) = 0    [ since we take only 3 cards ]

The joint probability distribution is :

		X
	f(X,Y)	0	1	2	3
Y	0	286/9139 = 22/703	1014/9139 = 78/703	1014/9139=78/703	286/9139=22/703
1	1014/9139 = 78/703	2197/9139 =169/703	1014/9139=78/703	0
2	1014/9139 = 78/703	1014/9139 = 78/703	0	0
3	286/9139 = 22/703	0	0	0

b) P[ (X,Y)$\epsilon$A] where A is the region given by [ (X,Y) | X+Y >= 2 ]

Here , P[ (X,Y)$\epsilon$A] = P(X+Y>=2)

= 1 - P(X+Y<2)

= 1 - [ P(0,0) + P(0,1) + P(1,0) ]

= 1 - [ 22/703 + 78/703 + 78/703 ]

= 1 - 178/703

= 525/703

$\therefore$ P[ (X,Y)$\epsilon$A] = 525/703