Solutions to HW3 Problem 2.2.1 N n PN - IUPUI

ECE302 Spring 2006

HW3 Solutions

February 2, 2006

1

Solutions to HW3

Note: Most of these solutions were generated by R. D. Yates and D. J. Goodman, the authors of our textbook. I have added comments in italics where I thought more detail was appropriate. I have also largely rewritten the solutions to problems 2.10.1, 2.10.2, and 2.10.3.

Problem 2.2.1 ?

The random variable N has PMF

PN (n) =

c(1/2)n n = 0, 1, 2,

0

otherwise.

(a) What is the value of the constant c?

(b) What is P [N 1]?

Problem 2.2.1 Solution

(a) We wish to find the value of c that makes the PMF sum up to one.

PN (n) =

c(1/2)n n = 0, 1, 2

0

otherwise

(1)

Therefore,

2 n=0

PN (n)

=

c

+

c/2

+

c/4

=

1,

implying

c

=

4/7.

(b) The probability that N 1 is

P [N 1] = P [N = 0] + P [N = 1] = 4/7 + 2/7 = 6/7

(2)

Problem 2.2.2 ?

For random variables X and R defined in Example 2.5, find PX (x) and PR(r). In addition, find the following probabilities:

(a) P [X = 0]

(b) P [X < 3]

(c) P [R > 1]

Problem 2.2.2 Solution

From Example 2.5, we can write the PMF of X and the PMF of R as

1/8 x = 0

3/8

x=1

1/4 r = 0

PX (x) = 3/8 x = 2

PR (r) = 3/4 r = 2

(1)

1/8

x=3

0

otherwise

0 otherwise

From the PMFs PX (x) and PR(r), we can calculate the requested probabilities

ECE302 Spring 2006

HW3 Solutions

February 2, 2006

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(a) P [X = 0] = PX (0) = 1/8. (b) P [X < 3] = PX (0) + PX (1) + PX (2) = 7/8. (c) P [R > 1] = PR(2) = 3/4.

Problem 2.2.3 ?

The random variable V has PMF

PV (v) =

cv2 v = 1, 2, 3, 4, 0 otherwise.

(a) Find the value of the constant c. (b) Find P [V {u2|u = 1, 2, 3, ? ? ?}].

(c) Find the probability that V is an even number.

(d) Find P [V > 2].

Problem 2.2.3 Solution

(a) We must choose c to make the PMF of V sum to one.

4

PV (v) = c(12 + 22 + 32 + 42) = 30c = 1

(1)

v=1

Hence c = 1/30.

(b) Let U = {u2|u = 1, 2, . . .} so that

P

[V

U ] = PV

(1) + PV

(4) =

1 30

+

42 30

=

17 30

(2)

(c) The probability that V is even is

P [V

is even] = PV

(2) + PV

(4) =

22 30

+

42 30

=

2 3

(3)

(d) The probability that V > 2 is

P

[V

>

2] =

PV

(3) + PV

(4)

=

32 30

+

42 30

=

5 6

(4)

ECE302 Spring 2006

HW3 Solutions

February 2, 2006

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Problem 2.2.4 ?

The random variable X has PMF

PX (x) =

c/x x = 2, 4, 8, 0 otherwise.

(a) What is the value of the constant c?

(b) What is P [X = 4]?

(c) What is P [X < 4]?

(d) What is P [3 X 9]?

Problem 2.2.4 Solution

(a) We choose c so that the PMF sums to one.

x

PX

(x)

=

c 2

+

c 4

+

c 8

=

7c 8

=

1

(1)

Thus c = 8/7.

(b)

P

[X

=

4]

=

PX

(4)

=

8 7?4

=

2 7

(2)

(c)

P

[X

<

4]

=

PX

(2)

=

8 7?2

=

4 7

(3)

(d)

P

[3

X

9]

=

PX

(4)

+

PX

(8)

=

8 7?4

+

8 7?8

=

3 7

(4)

Problem 2.3.1 ?

In a package of M&Ms, Y , the number of yellow M&Ms, is uniformly distributed between 5 and 15.

(a) What is the PMF of Y ?

(b) What is P [Y < 10]?

(c) What is P [Y > 12]?

(d) What is P [8 Y 12]?

ECE302 Spring 2006

HW3 Solutions

February 2, 2006

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Problem 2.3.1 Solution

(a) If it is indeed true that Y , the number of yellow M&M's in a package, is uniformly distributed between 5 and 15, then the PMF of Y , is

PY (y) =

1/11 y = 5, 6, 7, . . . , 15 0 otherwise

(1)

(b)

P [Y < 10] = PY (5) + PY (6) + ? ? ? + PY (9) = 5/11

(2)

(c)

P [Y > 12] = PY (13) + PY (14) + PY (15) = 3/11

(3)

(d)

P [8 Y 12] = PY (8) + PY (9) + ? ? ? + PY (12) = 5/11

(4)

Problem 2.3.4 ?

Anytime a child throws a Frisbee, the child's dog catches the Frisbee with probability p, independent of whether the Frisbee is caught on any previous throw. When the dog catches the Frisbee, it runs away with the Frisbee, never to be seen again. The child continues to throw the Frisbee until the dog catches it. Let X denote the number of times the Frisbee is thrown.

(a) What is the PMF PX (x)?

(b) If p = 0.2, what is the probability that the child will throw the Frisbee more than four times?

Problem 2.3.4 Solution

(a) Let X be the number of times the frisbee is thrown until the dog catches it and runs away. Each throw of the frisbee can be viewed as a Bernoulli trial in which a success occurs if the dog catches the frisbee an runs away. Thus, the experiment ends on the first success and X has the geometric PMF

PX (x) =

(1 - p)x-1p x = 1, 2, . . .

0

otherwise

(1)

(b) The child will throw the frisbee more than four times iff there are failures on the first 4 trials which has probability (1 - p)4. If p = 0.2, the probability of more than four throws is (0.8)4 = 0.4096.

ECE302 Spring 2006

HW3 Solutions

February 2, 2006

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Note: There is a less elegant but equally effective way to solve the problem, which I show below is equivalent.

P [X > 4] = 1 - (P [X = 4] + P [X = 3] + P [X = 2] + P [X = 1])

(2)

= 1 - p(1 - p)3 + p(1 - p)2 + p(1 - p) + p

(3)

= 1 - 4p + 6p2 - 4p3 + p4

(4)

= (1 - p)4

(5)

Problem 2.3.6 ?

The number of bits B in a fax transmission is a geometric (p = 2.5 ? 10-5) random variable. What is the probability P [B > 500,000] that a fax has over 500,000 bits?

Problem 2.3.6 Solution

The probability of more than 500,000 bits is

500,000

P [B > 500,000] = 1 -

PB (b)

(1)

b=1

500,000

=1-p

(1 - p)b-1

(2)

b=1

Math Fact B.4 implies that (1 - x)

500,000 b=1

xb-1

= 1 - x500,000.

Substituting,

x = 1 - p,

we

obtain:

P [B > 500,000] = 1 - (1 - (1 - p)500,000)

(3)

= (1 - 2.5 ? 10-5)500,000 0.3726 ? 10-5

(4)

Problem 2.3.7 ?

The number of buses that arrive at a bus stop in T minutes is a Poisson random variable B with expected value T /5.

(a) What is the PMF of B, the number of buses that arrive in T minutes?

(b) What is the probability that in a two-minute interval, three buses will arrive?

(c) What is the probability of no buses arriving in a 10-minute interval?

(d) How much time should you allow so that with probability 0.99 at least one bus arrives?

Problem 2.3.7 Solution

Since an average of T /5 buses arrive in an interval of T minutes, buses arrive at the bus stop at a rate of 1/5 buses per minute.

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