Solution to Exercise 5.06.

(a) The hint is that

$$P[ X=4 \, \& \, Y = 2 ] \; = \; P[ Y = 2 \vert X = 4 ] P [X = 4 ] .$$

Note that this follows from the definition of conditional probability:

$$P[ Y = 2 \vert X = 4 ] \; = \; \frac{P[ X=4 \, \& \, Y = 2 ]}{P [X = 4 ]} .$$

Now, if X = 4, that means 4 VCR's were purchased. We assume that each customer who buys a VCR is independent of the others and has a probability of 0.6 of buying the extended warranty. Thus, give X = 4, Y is a Bin(4,.6) random variable. Hence,

$$P[ Y = 2 \vert X = 4 ] \; = \; \left( \begin{array}{c} 4 \\ 2 \end{array} \right).6^{2} (1-.6)^2$$

$$\; = \; \frac{4*3}{1*2} * .36 * .16 \; = \; .3456$$

and hence

$$P[ X=4 \, \& \, Y = 2 ] \; = \; .3456*.25 = .0864 .$$

Note: this should be .3456*.15 = 0.05184. Sorry about the mistake. It's easier to correct it this way than put it in the equation.

(c) We will do part (c) next because we can use some of the results from part (c) for part (b). Of course always$Y \le X$, so we can start the table of the joint p.m.f.:
 
 

y=0 y=1 y=2 y=3 y=4

x=0 .1

x=1 .08 .12

$$\cdot \cdot \cdot$$ x=2

$$\cdot \cdot \cdot \cdot$$ x=3

$$\cdot \cdot \cdot \cdot \cdot$$ x=4

The entries for x=0 and x=1 are easy:

$$P[X=1 \, \& \, Y=1] \; = \; P[Y=1 \vert X=1] P[X=1] \; = \; .6 * .2 \; = \; .12$$

Now if X=2, Y has a (conditional) Bin(2,.6) distribution. Thus,

$$P[ Y = 0 \vert X = 2 ] \; = \; \left( \begin{array}{c} 2 \\ 0 \end{array} \right).6^{0} (1-.6)^2$$

$$\; = \; 1 * 1 * .16 \; = \; .16$$

and

$$P[ Y = 1 \vert X = 2 ] \; = \; \left( \begin{array}{c} 2 \\ 1 \end{array} \right).6^{1} (1-.6)^1$$

$$\; = \; 2 * .6 * .4 \; = \; .48$$

and finally

$$P[ Y = 2 \vert X = 2 ] \; = \; \left( \begin{array}{c} 2 \\ 2 \end{array} \right).6^{2} (1-.6)^0$$

$$\; = \; 1 * .36 * 1 \; = \; .36$$

Multiplying these by .3 = P[X=2] gives the nonzero entries in the third row as .48, .144, and .108.

We give the results for the next two rows without the detailed calculations:

$$P[Y=y\vert X=3] \; = \; \left\{ \begin{array}{ccl}0.064 & ... ...& \mbox{if } y=2 \\ 0.216 & & \mbox{if } y=3\end{array} \right.$$

These are multiplied by .25 = P[X=3] to obtain the nonzero entries in the fourth row.

$$P[Y=y\vert X=4] \; = \; \left\{ \begin{array}{ccl}0.0256 &... ... \mbox{if } y=3 \\ 0.1296 & & \mbox{if } y=4\end{array} \right.$$

These are multiplied by .15 = P[X=4] to obtain the nonzero entries in the fifth row.
 
 

y=0 y=1 y=2 y=3 y=4

x=0 .1

x=1 .08 .12

x=2 0.48 0.144 0.108

x=3 0.016 0.072 0.108 0.054

x=4 0.00384 0.02304 0.05184 0.05184 0.01944

sum 0.24784 0.35904 0.26784 0.10584 0.01944

We have summed the columns to obtain the marginal p.m.f. for Y, e.g. p_Y (2) $\; = \;$ 0.26784.

(b)

$$P[X=Y] \; = \; p_{XY} (0,0) +p_{XY} (1,1) +p_{XY} (2,2) +p_{XY} (3,3) +p_{XY} (4,4)$$

$$\; = \; 0.1 + 0.12 + 0.108 + 0.054 + 0.01944 \; = \; 0.40144 .$$

3/10/2001