Feb 28 Homework Solutions Math 151, Winter 2012 Chapter 6 ...

Feb 28 Homework Solutions Math 151, Winter 2012

Chapter 6 Problems (pages 287-291)

Problem 6 A bin of 5 transistors is known to contain 2 that are defective. The transistors are to be tested, one at a time, until the defective ones are identified. Denote by N1 the number of test made until the first defective is identified and by N2 the number of additional tests until the second defective is identified. Find the joint probability mass function of N1 and N2.

Note N1 and N2 take positive integer values such that N1 + N2 5, i.e. N1 = 1, . . . , 4 and N2 = 1, . . . , 5 - N1. Each pair of values for N1 and N2 are equally likely, and there are 10 such pairs. So

1 P {N1 = i, N2 = j} = 10 for i = 1, . . . , 4 and j = 1, . . . , 5 - i

Problem 8 The joint probability density function of X and Y is given by

f (x, y) = c(y2 - x2)e-y, -y x y, 0 < y < .

(a) Find c.

We want

y

1=

f (x, y)dxdy =

c(y2 - x2)e-ydxdy.

- -

0 -y

We compute

y

(y2 - x2)e-ydxdy =

y

(y2 - x2)e-ydxdy

0 -y

0 -y

=

xy2 - x3

x=y

e-y

dy

0

3

x=-y

4 =

y 3 e-y dy

30

= 8 e-ydy using integration by parts

0

= 8.

Thus c = 1/8.

(b) Find the marginal densities of X and Y .

1

The marginal probability density function of X is

fX(x) = f (x, y)dy

-

= 1 (y2 - x2)e-ydy |x| 8

= 1 ye-ydy using integration by parts |x| 4

= 1 |x|e-|x| + 1 e-ydy using integration by parts

4

|x| 4

= 1 |x|e-|x| + 1 e-|x|

4

4

= 1 e-|x| |x| + 1 4

Let fY be the marginal probability density function of Y . For y < 0 we have fY (y) = 0, and for y 0 we have

fY (y) = f (x, y)dx

-

1 =

y

(y2 - x2)e-ydx

8 -y

1 =

8

xy2 - x3 3

x=y

e-y

x=-y

= 1 y3e-y 6

(c) Find E[X].

E[X] =

1

xf (x, y)dxdy =

y

x(y2 - x2)e-ydxdy = 0,

- -

8 0 -y

since x(y2 - x2)e-y is an odd function of x.

Problem 10 The joint probability density function of X and Y is given by

f (x, y) = e-x-y, 0 x < , 0 y <

(a) Find P {X < Y }.

P {X < Y } = =

f (x, y)dydx =

e-x-ydydx =

0x

e-2xdx =

0

0x

-1 2

e-2x

x= x=0

=

1 .

2

-e-x-y

y= y=x

dx

0

2

(b) Find P {X < a}.

P {X < a} = =

a

a

f (x, y)dydx =

e-x-ydydx =

00

00

a

e-xdx =

-e-x

x=a x=0

= 1 - e-a.

0

a

-e-x-y y= dx y=0 0

Problem 15 The random vector (X, Y ) is said to be uniformly distributed over a region R in the plane if, for some constant c, its joint density is

f (x, y) =

c if (x, y) R, 0 otherwise.

(a) Show that 1/c = area of region R. Let R2 be the 2-dimensional plane. Observe that for E R2,

P {(X, Y ) E} = f (x, y)dxdy =

c = c ? area(E R).

E

ER

Since f (x, y) is a joint density function, we have

1 = P {(X, Y ) R2} = c ? area(R2 R) = c ? area(R).

So the area of R is 1/c.

(b) Suppose that (X, Y ) is uniformly distributed over the square centered at (0, 0) and with sides of length 2. Show that X and Y are independent, with each being distributed uniformly over (-1, 1).

Let R be the real line. For sets A, B R, we have

1

P {X A, Y B} =

f (x, y)dxdy =

dxdy

BA

B[-1,1] A[-1,1] 4

1 = length(A [-1, 1])length(B [-1, 1])

4

and

1

1

1

P {X A} =

f (x, y)dxdy =

dxdy = length(A [-1, 1])

- A

-1 A[-1,1] 4

2

and

11

1

P {Y B} =

f (x, y)dxdy =

dxdy = length(B [-1, 1]).

B -

B[-1,1] -1 4

2

Thus

1 P {X A, Y B} = length(A[-1, 1])length(B[-1, 1]) = P {X A}P {Y B},

4

so X and Y are independent. Also, our formulas for P {X A} and P {Y B} show that each of X and Y are uniformly distributed over (-1, 1).

3

(c) What is the probability that (X, Y ) lies in the circle of radius 1 centered at the origin? That is, find P {X2 + Y 2 1}.

Let C denote the interior of the circle of radius 1 centered at the origin. We want to compute P {(X, Y ) C}. We have

1

1

1

P {(X, Y ) C} = ? area(C R) = ? area(C) = ? = .

4

4

4

4

Problem 16 Suppose that n points are independently chosen at random on the circumference of a circle, and we want the probability that they all lie in some semicircle. That is, we want the probability that there is a line passing through the center of the circle such that all the points are on one side of that line. Let P1, . . . , Pn denote the n points. Let A denote the event that all the points are contained in some semicircle, and let Ai be the event that all the points lie in the semicircle beginning at the point Pi and going clockwise for 180, i = 1, . . . , n.

(a) Express A in terms of the Ai.

It's clear that Ai A for each i. Hence

n i=1

Ai

A.

Now we want to show the

reverse inclusion. If A occurs, then we can rotate the semicircle clockwise until it

starts at one of the points Pi, showing that Ai and thus

n i=1

Ai

occurs.

Hence

A

n i=1

Ai.

So

we've

shown

that

A

=

n i=1

Ai.

(b) Are the Ai mutually exclusive?

Almost, but not quite. Strictly speaking the Ai and Aj are not mutually exclusive since if Pi = Pj, the other Pk can lie in the semicircle beginning at Pi = Pj and going clockwise for 180 degrees in which case both Ai and Aj occur. However, if Pi = Pj, then Ai and Aj cannot both occur since if Pj is in the semicircle starting at Pi and going clockwise 180 degrees, then Pi is not in the semicircle starting at Pj and going clockwise 180 degrees. So AiAj {Pi = Pj}. Moreover P {Pi = Pj} = 0, so P (AiAj) = 0.

(c) Find P (A).

We argued in part (b) that P (AiAj) = 0. It follows that P (AiAjAk) = 0, and the same is true for all higher order intersections.

We have

n

P (A) = P ( Ai)

i=1 n

= P (Ai) - P (AiAj) +

P (AiAjAk) - . . .

i=1

i L/3. Alternatively, after drawing this region in the plane, you can compare the

area

of

this

region

to

the

area

of

the

larger

rectangle

given

by

0

x

L 2

and

L 2

y

L

to see that the desired probability is 7/9.

5

 ................
................

In order to avoid copyright disputes, this page is only a partial summary.

Google Online Preview   Download