7 A refrigerator manufacturer subjects his finished products

7. A refrigerator manufacturer subjects his finished products to a final inspection. Of
interest are two categories of defects: scratches or flaws in the porcelain finish, and
mechanical defects. The number of each type of defect is a random variable. The results
of inspecting 40 are shown in the following table, where X represents the occurrence of
finish defects and Y represents the occurrence of mechanical defects.
Y
X
0 1 2 3 4 5
0 1/40 3/40 2/40 2/40 4/40 3/40
1 5/40 2/40 2/40 1/40 1/40 0
2 4/40 2/40 2/40 1/40 0 0
3 3/40 1/40 0 0 0 0
4 1/40 0 0 0 0 0
(a) Find the marginal distributions of X and Y.
(b) Find the probability distribution of mechanical defects given that there are no finish
defects.
(c) Probability that X>3.
(d) Probability that Y<= 2.

Solution

a) here X takes the values 0,1,2,3,4            Y takes the values 0,1,2,3,4,5

P[X=0]=P[X=0,Y=0]+P[X=0,Y=1]+P[X=0,Y=2]+P[X=0,Y=3]+P[X=0,Y=4]+P[X=0,Y=5]

          =1/40+3/40+2/40+2/40+4/40+3/40=15/40

P[X=1]=P[X=1,Y=0]+P[X=1,Y=1]+P[X=1,Y=2]+P[X=1,Y=3]+P[X=1,Y=4]+P[X=1,Y=5]

          =5/40+2/40+2/40+1/40+1/40+0=11/40

P[X=2]=P[X=2,Y=0]+P[X=2,Y=1]+P[X=2,Y=2]+P[X=2,Y=3]+P[X=2,Y=4]+P[X=2,Y=5]

          =4/40+2/40+2/40+1/40+0+0=9/40

P[X=3]=P[X=3,Y=0]+P[X=3,Y=1]+P[X=3,Y=2]+P[X=3,Y=3]+P[X=3,Y=4]+P[X=3,Y=5]

          =3/40+1/40+0+0+0+0=4/40

P[X=4]=P[X=4,Y=0]+P[X=4,Y=1]+P[X=4,Y=2]+P[X=4,Y=3]+P[X=4,Y=4]+P[X=4,Y=5]

          =1/40+0+0+0+0+0=1/40

so marginal of X is

X:               0                   1                       2                      3                         4

P[X=x]=          15/40             11/40                 9/40                  4/40                    1/40

now for Y

P[Y=0]=P[X=0,Y=0]+P[X=1,Y=0]+P[X=2,Y=0]+P[X=3,Y=0]+P[X=4,Y=0]

          =1/40+5/40+4/40+3/40+1/40=14/40

P[Y=1]=P[X=0,Y=1]+P[X=1,Y=1]+P[X=2,Y=1]+P[X=3,Y=1]+P[X=4,Y=1]

          =3/40+2/40+2/40+1/40+0=8/40

P[Y=2]=P[X=0,Y=2]+P[X=1,Y=2]+P[X=2,Y=2]+P[X=3,Y=2]+P[X=4,Y=2]

          =2/40+2/40+2/40+0+0=6/40

P[Y=3]=P[X=0,Y=3]+P[X=1,Y=3]+P[X=2,Y=3]+P[X=3,Y=3]+P[X=4,Y=3]

          =2/40+1/40+1/40+0+0=4/40

P[Y=4]=P[X=0,Y=4]+P[X=1,Y=4]+P[X=2,Y=4]+P[X=3,Y=4]+P[X=4,Y=4]

          =4/40+1/40+0+0+0=5/40

P[Y=5]=P[X=0,Y=5]+P[X=1,Y=5]+P[X=2,Y=5]+P[X=3,Y=5]+P[X=4,Y=5]

          =3/40+0+0+0+0=3/40

so marginal of Y is

Y:                0                  1                  2                   3                    4                         5

P[Y=y]:       14/40            8/40             6/40               4/40                5/40                   3/40

b) we need to find the probability distribution of Y|X=0

so P[Y=0|X=0]=P[X=0,Y=0]/P[X=0]=(1/40)/(15/40)=1/15

P[Y=1|X=0]=P[X=0,Y=1]/P[X=0]=(3/40)/(15/40)=3/15

P[Y=2|X=0]=P[X=0,Y=2]/P[X=0]=(2/40)/(15/40)=2/15

P[Y=3|X=0]=P[X=0,Y=3]/P[X=0]=(2/40)/(15/40)=2/15

P[Y=4|X=0]=P[X=0,Y=4]/P[X=0]=(4/40)/(15/40)=4/15

P[Y=5|X=0]=P[X=0,Y=5]/P[X=0]=(3/40)/(15/40)=3/15

so the distribution of Y|X=0 is

Y|X=0            0          1                2                3                  4                   5

P[Y=y|X=0]:        1/15             3/15            2/15           2/15              4/15              3/15

c) P[X>3]=P[X=4]=1/40

d) P[Y<=2]=P[Y=0]+P[Y=1]+P[Y=2]=14/40+8/40+6/40=28/40=7/10