# EE 562a Random Processes in Engineering

**Topics:**Probability theory, Conditional probability, Probability

**Pages:**6 (1583 words)

**Published:**September 16, 2014

EE department, USC, Fall 2014

Instructor: Prof. Salman Avestimehr

Homework 1

Solutions

1. (Axioms of Probability) Prove the union bound:

n

P [∪n Ak ] ≤

k=1

P [Aj ].

j=1

The union bound is useful because it does not require that the events Aj be independent or disjoint.

Problem 1 Solution

We prove this part by induction, for k = 2 we have

P (A1 ∪ A2 ) = P (A1 ) + P (A2 ) − P (A1 ∩ A2 ) ≤ P (A1 ) + P (A2 )

(1)

Now, assume that the statement is true for k = n

n

P (A1 ∪ A2 . . . ∪ An ) ≤

P (Ai )

(2)

i=1

For k = n + 1, we can write

P (A1 ∪ A2 . . . ∪ An+1 ) = P (An+1 ) + P (A1 ∪ A2 . . . ∪ An ) − P (An+1 ∩ (A1 ∪ A2 . . . ∪ An )) n+1

≤

P (Ai )

i=1

(3)

This proves the union bound.

2. (Independence) For each one of the following statements, indicate whether it is true or false, and provide a brief explanation.

(a) If P (A|B) = P (A), then P (B|Ac ) = P (B).

(b) If 5 out 10 independent fair coin tosses resulted in tails, the events “ﬁrst toss was tails” and “10th toss was tails” are independent.

(c) If 10 out 10 independent fair coin tosses resulted in tails, the events “ﬁrst toss was tails” and “10th toss was tails” are independent.

Problem 2 Solution

(a) True.

P (A|B) = P (A), therefore,

P (A ∩ B) = P (A)P (B)

(4)

But we know that (A∩B) and (Ac ∩B) are disjoint events, and (A∩B)∪(Ac ∩B) = B. Therefore,

P (A ∩ B) + P (Ac ∩ B) = P (B)

(5)

Thus, from (4) and (5) we can write

=⇒

=⇒

P (B) − P (Ac ∩ B) = P (A)P (B)

P (Ac ∩ B) = (1 − P (A))P (B) = P (Ac )P (B)

P (Ac )P (B)

= P (B)

P (B|Ac ) =

P (Ac )

(b) False.

Let A designate the event “5 out of 10 independent fair coin tosses are tails”. Also, let B designate the event “ﬁrst toss of the fair coin results in tails”. In addition, let C designate the event “10th toss of the fair coin results in tails”. We have

P (B|A) =

1

2

P (C|A) =

But,

P (A ∩ B)

=

P (A)

P (A ∩ C)

=

P (A)

1

2

∗

9

4

∗

9

4

10

5

10

5

∗ ( 1 )9

2

=

∗ ( 1 )9

2

=

1

∗ ( 2 )10

1

∗ ( 2 )10

9

4

10

5

9

4

10

5

1

( 1 )2 ∗ 8 ∗ ( 2 )8

P (A ∩ B ∩ C)

2

3

P (B ∩ C|A) =

=

=

10

P (A)

∗ ( 1 )10

5

2

8

3

10

5

Therefore, P (B ∩ C|A) = P (B|A)P (C|A). So (B|A) is not independent of (C|A).

(c) True.

Let Let A designate the event “10 out of 10 independent fair coin tosses are tails”. Also, let B designate the event “ﬁrst toss of the fair coin results in tails”. In addition, let C designate the event “10th toss of the fair coin results in tails”. We have

P (B|A) = 1

P (C|A) = 1

P (B ∩ C|A) = 1

Therefore, P (B ∩ C|A) = P (B|A)P (C|A). Thus, (B|A) and (C|A) are independent. 2

3. (Conditional Probability) Each computer chip produced at a factory has a probability d of being defective (D), independent of other chips.

(a) What is the probability of ﬁnding k defective chips in a sample of n chips? (b) Now suppose that chips are tested in sequence, one by one, until a total of k defective chips are discovered. Find the probability that the kth defective chip is found after exactly n chips are sampled.

Now suppose that the testing procedure is unreliable:

Missed Detection: P [test says OK | chip is Defective] = t

False Alarm: P [test says Defective | chip is OK] = t, for some t ∈ (0, 1). (c) Suppose that each chip is tested 10 times independently, and rejected if at least 6 of the tests come up defective. Given that a chip is rejected, compute the probability that it is defective as a function of t and d.

Problem 3 Solution

(a)

(b)

n k

d (1 − d)n−k

k

n−1 k

d (1 − d)n−k .

k−1

This is the probability that the nth sample is defective and from the ﬁrst n − 1 samples exactly k − 1 of them are defective.

(c) Let A designate the event “at least 6 out of 10 independent tests come up defective”. Also, let D designate the event “the chip is actually defective”. Therefore,

(D|A) designates the event “A...

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