Question

Part 1 Suppose that 2 batteries are randomly chosen without replacement from a group of 12 batteries: 3 new, 4 used (working), and 5 defective. Let the random variable X denote the number of new batteries chosen and the random variable Y denote the number of used batteries chosen. The joint distribution fxy is given in the following table: 0 12 17663/6 120/6612/66 1. Calculate P ( X 1 ,Y > 1) 2. Find the marginal probability mass function fx (x) of the random Variable X by completing the following table and calculate its mean.

Answer 1

1.

Calculate P(X$\small \leq$1, Y$\small \geq$1)

X$\small \leq$1 : X=0, X=1

Y$\small \geq$1 : Y=1, Y=2

$\small \\P(X\leq 1,Y\geq 1)\\\\= P(X=0,Y=1)+P(X=0,Y=2) + P(X=1,Y=1)+P(X=1,Y=2)$

From the table

P(X=0,Y=1) = 20/66 ; P(X=0,Y=2) = 6/66 ; P(X=1,Y=1) = 12/66 ; P(X=1,Y=2) = 0

$\small \\P(X\leq 1,Y\geq 1)\\\\= P(X=0,Y=1)+P(X=0,Y=2) + P(X=1,Y=1)+P(X=1,Y=2)\\\\=20/66 +6/66+12/66+0=38/66=19/33$

$\small \\P(X\leq 1,Y\geq 1)=38/66=19/33$

2.

Marginal probability mass function fX(x) of the random Variable X

$\small f_{X}(x)=\sum_{y }^{ }f_{XY}(x,y)$

For X=0

$\small \\f_{X}(0)=\sum_{y=0 }^{ 2}f_{XY}(0,y) \\\\= f_{XY}(0,0)+f_{XY}(0,1)+f_{XY}(0,2)=10/66+20/66+6/66=36/66$

$\small \\f_{X}(1)=\sum_{y=0 }^{ 2}f_{XY}(1,y) \\\\= f_{XY}(1,0)+f_{XY}(1,1)+f_{XY}(1,2)=15/66+12/66+0=27/66$

$\small \\f_{X}(2)=\sum_{y=0 }^{ 2}f_{XY}(2,y) \\\\= f_{XY}(2,0)+f_{XY}(2,1)+f_{XY}(2,2)=3/66+0+0=3/66$

x	0	1	2
fX(x)	36/66	27/66	3/66

or

x	0	1	2
fX(x)	12/22	9/22	1/22

Mean : E(X)

$\small \\E(X)=\sum_{x=0}^{2}xf_{X}(x)=0\times f_{X}(0)+1\times f_{X}(1)+2\times f_{X}(2)\\\\\\=0\times 12/22+1\times 9/22+2\times 1/22=0+9/22+2/22=11/22=1/2$

Mean = 1/2 =0.5