Bulbs. Refer to Exercise 5.43. Assume that the number of defective bulbs on different boxes are independent of one another. Let X and Y denote the number of defective bulbs on each of two boxes.

x y

0 1 2 P(X = x)

0 1

2 P( Y = y)

a. Complete the preceding joint probability distribution table. Hint:

To obtain the joint probability in the first row, third column, use the definition of independence for discrete random variables and the table in Exercise 5.43:

P({X = 0} & {Y = 2}) = P(X = 0) · P(Y = 2)

= 0.181 · 0.36

= 0.065.

b. Use the joint probability distribution you obtained in part

(a) to determine the probability distribution of the random variable X + Y , the total number of defective bulbs in two boxes; that is, complete the following table.

u 01234 P(X + Y) = u

c. Use part

(b) to find μX+Y and σ2 X+Y .

d. Use part

(c) to verify that the following equations hold for this example:

μX+Y = μX + μY σX+Y = σX + σY .

(Note: The mean and variance of X and Y are the same as that of X in Exercise 5.43.)

e. The equations in part

(d) hold in general: If X and Y are any two random variables,

μX+Y = μX + μY .

In addition, if X and Y are independent,

σ2 X+Y = σ 2 Y + σ 2 X .

Interpret these two equations in words.