2.1 LET 𝑨, 𝑩 AND 𝑪 BE ARBITRARY SETS. DETERMINE WHICH OF THE FOLLOWING

RELATIONS ARE CORRECT AND WHICH ARE INCORRECT:

a) 𝑨𝑩(𝑪 ∪ 𝑩) = 𝐴𝐵𝐶 𝑈 𝐴𝐵𝐵 = 𝐴𝐵𝐶 𝑈 𝐴𝐵 ≠ 𝑨𝑩𝑪 Incorrecto

b) 𝑨𝑩̅̅ ̅̅ = 𝐴 ∩ 𝐵̅̅ ̅̅ ̅̅ ̅̅ ̅ = �̅� ∪ �̅� (Ley de Morgan) Correcto

c) 𝑨 ∪ 𝑩̅̅ ̅̅ ̅̅ ̅̅ ̅ = �̅� ∩ �̅� = 𝑨 ̅𝑩 ̅ (Ley de Morgan) Correcto

d) (𝑨 ∪ 𝑩)̅̅ ̅̅ ̅̅ ̅̅ ̅̅ ̅̅ 𝑪 = (�̅� ∩ �̅�)𝐶 = 𝑨 ̅𝑩 ̅ 𝑪 Correcto

2.4 LET𝑺 = {𝟏, 𝟐, … , 𝟏𝟎}, 𝑨 = {𝟏, 𝟑, 𝟓}, 𝑩 = {𝟏, 𝟒, 𝟔}, And 𝑪 = {𝟐, 𝟓, 𝟕}. DETERMINE

ELEMENTS OF THE FOLLOWING SETS:

a) 𝑺 ∪ 𝑪 = {1,2,3,4,5,6,7,8,9,10}

b) 𝑨 𝑼 𝑩 = {1,3,4,5,6}

c) �̅�𝑪

�̅� = {2,4,6,7,8,9,10} → �̅�𝐶 = {2,7}

d) 𝑨 ̅ ∪ (𝑩𝑪)

𝐵𝐶 = {∅} → 𝐴 ̅ ∪ (𝐵𝐶) = {2,4,6,7,8,9,10}

e) 𝑨𝑩𝑪 ̅̅ ̅̅ ̅̅ ̅

𝐴𝐵𝐶 = {∅} → 𝐴𝐵𝐶 ̅̅ ̅̅ ̅̅ ̅ = {1,2,3,4,5,6,7,8,9,10}

f) 𝑨 ̅𝑩 ̅̅̅ ̅̅ ̅̅

�̅� = {2,4,6,7,8,9,10} , �̅� = {2,3,5,7,8,9,10}

𝐴 ̅𝐵 ̅ = {2,7,8,9,10} → 𝐴 ̅𝐵 ̅̅̅ ̅̅ ̅̅ = {1,3,4,5,6}

g) (𝑨𝑩) ∪ (𝑩𝑪) ∪ (𝑪𝑨)

𝐴𝐵 = {1}, 𝐵𝐶 = {∅}, 𝐶𝐴 = {5} → (𝐴𝐵) ∪ (𝐵𝐶) ∪ (𝐶𝐴) = {1,5}

2.5 REPEAT PROBLEM 2.4 if 𝑺 = {𝒙: 𝟎 ≤ 𝒙 ≤ 𝟏𝟎}, 𝑨 = {𝒙: 𝟏 ≤ 𝒙 ≤ 𝟓}, 𝑩 = {𝒙: 𝟏 ≤ 𝒙 ≤

𝟔}, AND 𝑪 = {𝒙: 𝟐 ≤ 𝒙 ≤ 𝟕}

a) 𝑺 ∪ 𝑪 = {𝑥: 0 ≤ 𝑥 ≤ 10}

b) 𝑨 ∪ 𝑩 = {𝑥: 1 ≤ 𝑥 ≤ 6}

c) �̅�𝑪

�̅� = {𝑥: 0 ≤ 𝑥 < 1, 5 < 𝑥 ≤ 10 } → �̅�𝐶 = {𝑥: 5 < 𝑥 ≤ 7}

d) 𝑨 ̅ ∪ (𝑩𝑪)

𝐵𝐶 = {𝑥: 2 ≤ 𝑥 ≤ 6} → 𝐴 ̅ ∪ (𝐵𝐶) = {𝑥: 0 ≤ 𝑥 < 1, 2 ≤ 𝑥 ≤ 10}

e) 𝑨𝑩𝑪 ̅̅ ̅̅ ̅̅ ̅

𝐴𝐵𝐶 = {𝑥: 2 ≤ 𝑥 ≤ 5} → 𝐴𝐵𝐶 ̅̅ ̅̅ ̅̅ ̅ = {𝑥: 0 ≤ 𝑥 < 2, 5 < 𝑥 ≤ 10}

f) 𝑨 ̅𝑩 ̅̅̅ ̅̅ ̅̅

�̅� = {𝑥: 0 ≤ 𝑥 < 1, 5 < 𝑥 ≤ 10 } , �̅� = {𝑥: 0 ≤ 𝑥 < 1, 6 < 𝑥 ≤ 10 }

𝐴 ̅𝐵 ̅ = {𝑥: 0 ≤ 𝑥 < 1, 6 < 𝑥 ≤ 10 } → 𝐴 ̅𝐵 ̅̅̅ ̅̅ ̅̅ = {𝑥: 1 ≤ 𝑥 ≤ 6 }

g) (𝑨𝑩) ∪ (𝑩𝑪) ∪ (𝑪𝑨)

𝐴𝐵 = {𝑥: 1 ≤ 𝑥 ≤ 5}, 𝐵𝐶 = {𝑥: 2 ≤ 𝑥 ≤ 6}, 𝐶𝐴 = {𝑥: 2 ≤ 𝑥 ≤ 5}

(𝐴𝐵) ∪ (𝐵𝐶) ∪ (𝐶𝐴) = {𝑥: 1 ≤ 𝑥 ≤ 6}

2.6 DRAWN VENN DIAGRAMS OF EVENTS A AND B REPRESENTING THE FOLLOWING

SITUATIONS:

a) 𝐴 and 𝐵 are arbitrary

b) If 𝐴 occurs, 𝐵 must occur

c) If 𝐴 occurs, 𝐵 cannot occur

d) 𝐴 and 𝐵 are independent.

2.9 AN ENGINEERING SYSTEM HAS TWO COMPONENTS. LET US DEFINE THE FOLLOWING

EVENTS:

𝐴: First component is good. �̅�: First component is defective.

𝐵: Second component is good. �̅�: Second component is defective.

Describe the following events in terms of 𝐴, �̅�, 𝐵, 𝑎𝑛𝑑 �̅�:

a) At least one of the components is good 𝐴 ∪ 𝐵

b) One is good and one is defective 𝐴�̅� ∪ �̅�𝐵

2.11 A SATELLITE CAN FAIL FOR MANY POSSIBLE REASONS, TWO OF WHICH ARE COMPUTER

FAILURE AND ENGINE FAILURE. FOR A GIVEN MISSION, IT IS KNOWN THAT:

The probability of engine failure is 0.008

The probability of computer failure is 0.001

Given engine failure, the probability of satellite failure is 0.98

Given computer failure, the probability of satellite failure is 0.45

Given any other component failure, the probability of satellite failure is zero.

a) Determine the probability that a satellite fails.

𝑃(𝑓𝑡) = 0.008 ∗ 0.98 + 0.001 ∗ 0.45 = 8.29 ∗ 10−3

b) Determine the probability that a satellite fails and is due to engine failure.

𝑃(𝑓𝑚𝑜𝑡𝑜𝑟) = 0.008 ∗ 0.98 = 7.84 ∗ 10−3

c) Assume that engines in different satellites perform independently. Given a satellite has

failed as result of engine failure, what is the probability that the same will happen to

another satellite?

𝑃(𝑓𝑚𝑜𝑡𝑜𝑟 𝑠𝑎𝑡𝑒𝑙𝑖𝑡𝑒𝑠 𝑖𝑛𝑑𝑒𝑝𝑒𝑛𝑑𝑖𝑒𝑛𝑡𝑒𝑠) = 0.008 ∗ 0.98 = 7.84 ∗ 10−3

2.14 A BOX CONTAINS 20 PARTS, OF WHICH 5 ARE DEFECTIVE. TWO PARTS ARE DRAWN AT

RANDOM FROM THE BOX. WHAT IS THE PROBABILITY THAT:

a) Both are good?

𝑃 =15

20∗

14

19= 0.553

b) Both are defective?

𝑃 =5

20∗

4

19= 0.053

c) One is good and one is defective?

𝑃 =15

20∗

5

19+

5

20∗

15

19= 0.395

2.15 AN AUTOMOBILE BRAKING DEVICE CONSISTS OF THREE SUBSYSTEMS, ALL OF WHICH

MUST WORK FOR THE DEVICE TO WORK. THESE SYSTEMS ARE AN ELECTRONIC SYSTEM, A

HYDRAULIC SYSTEM, AND A MECHANICAL ACTIVATOR. IN BRAKING, THE RELIABILITIES

(PROBABILITIES OF SUCCESS) OF THESE UNITS ARE 0.96, 0.95 AND 0.95 RESPECTIVELY.

ESTIMATE THE SISTEM RELIABILITY ASSUMING THAT THESE SUBSYSTEMS FUNCTION

INDEPENDENTLY.

Comment: systems of this type can be graphically represented as shown in Figure 2.10,

in which subsystems 𝐴 (electronic system), 𝐵 (hydraulic system), and 𝐶(mechanical

activator) are arranged in series. Consider the path 𝑎 → 𝑏 as the ‘path to success’. A

breakdown of any or all of 𝐴, 𝐵, or 𝐶 will block the path from 𝑎 to 𝑏.

𝑅 = 𝑃(𝐴) ∗ 𝑃(𝐵) ∗ 𝑃(𝐶) = 0.96 ∗ 0.95 ∗ 0.95 = 0.8664

2.16 A SPACECRAFT HAS 1000 COMPONENTS IN SERIES. IF THE REQUIRED RELIABILITY OF

THE SPACECRAFT IS 𝟎. 𝟗 AND IF ALL COMPONENTS FUNCTION INDEPENDENTLY AND HAVE

THE SAME RELIABILITY, WHAT IS THE REQUIRED RELIABILITY OF EACH COMPONENT?

𝑅 = 0. 91000−1= 0.999895

2.17 AUTOMOBILES ARE EQUIPPED WITH REDUNDANT BRAKING CIRCUITS; THEIR BRAKES

FAIL ONLY WHEN ALL CIRCUITS FAIL. CONSIDER ONE WITH TWO REDUNDANT BRAKING

CIRCUITS, EACH HAVING A RELIABILITY OF 𝟎. 𝟗𝟓. DETERMINE THE SYSTEM RELIABILITY

ASSUMING THAT THESE CIRCUITS ACT INDEPENDENTLY.

Comment: systems of this type are graphically represented as in Figure 2.11, in which

the circuits (𝐴 and 𝐵) have a parallel arrangement. The path to success is broken only

when breakdowns of both 𝐴 and 𝐵 occur.

𝑃(𝐴 ∪ 𝐵) = 1 − (1 − 0.95)(1 − 0.95) = 0.998

2.18 ON THE BASIS OF THE DEFINITIONS GIVEN IN PROBLEMS 2.15 AND 2.17 FOR SERIES AND

PARALLEL ARRANGEMENTS OF SYSTEM COMPONENTS, DETERMINE RELIABILITIES OF THE

SYSTEMS DESCRIBED BY THE BLOCK DIAGRAMS AS FOLLOWS.

a) The diagram

𝑃(𝐵 ∪ 𝐶) = 1 − (1 − 0.85)(1 − 0.90) = 0.985

b) The diagram

𝑃(𝐴 ∩ (𝐵 ∪ 𝐶)) = 0.90 ∗ 0.985 = 0.887

2.20 EVENTS 𝑨 AND 𝑩 ARE MUTUALLY EXCLUSIVE. CAN THEY ALSO BE INDEPENDENT?

EXPLAIN.

No, ya que dos eventos mutuamente exclusivos no pueden presentarse a la misma vez,

a diferencia de dos eventos independientes ya que estos pueden o no ocurrir al mismo

tiempo.

2.21 LET 𝑷(𝑨) = 𝟎. 𝟒, AND 𝑷(𝑨 ∪ 𝑩) = 𝟎. 𝟕. WHAT IS 𝑷(𝑩) IF:

a) 𝐴 and 𝐵 are independent?

𝑃(𝐴 ∪ 𝐵) = 𝑃(𝐴) + 𝑃(𝐵) − 𝑃(𝐴) ∗ 𝑃(𝐵) ⇒ 0.7 = 0.4 + 𝑃(𝐵) − 0.4 ∗ 𝑃(𝐵) ⇒ 𝑃(𝐵) = 0.5

b) 𝐴 and 𝐵 are mutually exclusive.

𝑃(𝐴 ∪ 𝐵) = 𝑃(𝐴) + 𝑃(𝐵) ⇒ 0.7 = 0.4 + 𝑃(𝐵) ⇒ 𝑃(𝐵) = 0.7 − 0.4 = 0.3

2.22 LET 𝑷(𝑨 ∪ 𝑩) = 𝟎. 𝟕𝟓, AND 𝑷(𝑨𝑩) = 𝟎. 𝟐𝟓. IS IT POSSIBLE TO DETERMINE 𝑷(𝑨)

AND 𝑷(𝑩)?

Answer the same question if, in addition:

a) 𝐴 and 𝐵 are independent?

𝑃(𝐴 ∪ 𝐵) = 𝑃(𝐴) + 𝑃(𝐵) − 𝑃(𝐴)𝑃(𝐵) ⇒ 0.75 = 𝑃(𝐴) + 𝑃(𝐵) − 0.25 ⇒ 𝑃(𝐴) + 𝑃(𝐵)

= 1

También se tiene que: 𝑃(𝐴) ∗ 𝑃(𝐵) = 0.25

Por lo que: 𝑃(𝐴) =0.25

𝑃(𝐵)

0.25

𝑃(𝐵)+ 𝑃(𝐵) = 1 ⇒ 0.25 + [𝑃(𝐵)]2 = 𝑃(𝐵) ⇒ [𝑃(𝐵)]2 − 𝑃(𝐵) + 0.25 = 0

⇒ 𝑃(𝐴) = 0.5, 𝑃(𝐵) = 0.5

b) 𝐴 and 𝐵 are mutually exclusive.

𝑃(𝐴 ∪ 𝐵) = 𝑃(𝐴) + 𝑃(𝐵)

Debido a la falta de información no se puede resolver el ejercicio.