Lebanese American University School of Engineering and Architecture Department of Electrical, Computer, Industrial & Mechanical Engineering

KEY #1

Fall 2003 Page 1

COURSE : GNE 331, Probability & Statistics DATE GIVEN : Wednesday October 8, 2003 DUE DATE : Wednesday October 15, 2003 SEMESTER : Fall 2003 PROFESSOR : Dr. Raymond Ghajar

Probability

15. Consider the sample space

S = {copper, sodium, nitrogen, potassium, uranium, oxygen, zinc}

and the events

A = {copper, sodium, zinc}, B = {sodium, nitrogen, potassium}, C = {oxygen}.

List the elements of the sets corresponding to the following events:

(a) A′ (b) CA∪ (c) ( ) CBA ′∪′∩ (d) CB ′∩′ (e) CBA ∩∩ (f) ( ) ( )CABA ∪′∩′∪′

P

15. This problem can be visualized using Venn diagram.

coppernitrogen

potassium

uraniumoxygen

zincA

BC

S

(a) A’ = {nitrogen, potassium, uranium, oxygen}

(b) A∪C = {copper, sodium, oxygen, zinc}

(c) (A∩B’)∪C’ = {copper, sodium, nitrogen, potassium, uranium, zinc}

(d) B’∩C’ = {copper, uranium, zinc}

(e) A∩B∩C = φ

(f) (A’∪B’)∩ (A’∩C) = {oxygen}

S

17. Let A, B and C be events relative to the sample space S. Using Venn P

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diagrams, shade the areas representing the following events:

(a) ( )′∩ BA

(b) ( )′∪ BA (c) ( ) BCA ∪∩

17. (a) ( )′∩ BA

A B

S

(b) ( )′∪ BA

A B

S

(c) ( ) BCA ∪∩

A

S

B

C

S

5. A certain shoe comes in 5 different styles with each style available in 4 distinct colors. If the store wishes to display pairs of these shoes showing all of its various styles and colors, how many different pairs would the store have on display?

P

5. With n1 = 5 different styles of shoes and n2 = 4 distinct colors, the generalized multiplication rule yields:

N = n1*n2 = 5*4 = 20 different ways of displaying the shoes

S

7. A developer of a new subdivision offers a prospective home buyer a choice of 4 designs, 3 different heating systems, a garage or carport, and a patio or screened porch. How many different plans are available to this buyer?

P

7. There are n1 = 4 designs S

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n2 = 3 heating systems n3 = 2 types of car parking n4 = 2 types of patio/porch

Therefore, there are n1*n2*n3*n4 = 4*3*2*2 = 48 different plans

14. (a) In how many ways can 6 people be lined up to get on a bus?

(b) If 3 specific persons insist on following each other, how many ways are possible?

(c) If 2 specific persons refuse to follow each other, how many ways are possible?

P

14. (a) Six people can be arranged in 6! = 720 ways.

(b) Three specific persons can follow each other in 3! = 6 ways. If placed in a line with three other people (4 distinct), there will be 4P1 = 4!/(4-1)! = 4 different ways of placing them and 3! = 6 ways of placing the other three people. Therefore, there are 6*4*6 = 144 ways

(c) Two specific persons can follow each other in 2! = 2 ways. If placed in a line with four other people (5 distinct), there will be 5P1 = 5!/(5-1)! = 5 different ways of placing them and 4! = 24 ways of placing the other three people. Therefore, there are 2*5*24 = 240 ways to line up 6 people with a certain 2 of them together. Since there are a total of 720 ways of lining up all 6 people, there will be 720 – 240 = 480 ways of lining them up if a certain 2 of them refuse to follow each other.

S

25. How many distinct permutations can be made from the letters of the word infinity?

P

25. There are n = 8 letters in the word infinity, of which there are:

n1 = 3 i’s, n2 = 2 n’s, n3 = 1 f, n4 = 1 t, and n5 = 1 y.

The number of distinct permutations is given by:

( )( )( )( )( ) 3360!1!1!1!2!3

!8!!!!!

!,,,, 5432154321

===

nnnnn

nnnnnn

n

S

28. Nine people are going on a skiing trip in 3 cars that hold 2, 4, and 5 passengers respectively. In how many ways is it possible to transport the 9 people to the ski lodge, using all cars?

P

28. The 9 people need to be partitioned over the three cars as follows:

44103,4,2

94,3,2

95,2,2

94,4,1

95,3,1

9=

+

+

+

+

S

3. A box contains 500 envelopes of which 75 contain $100 in cash, 150 contain P

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$25, and 275 contain $10. An envelope may be purchased for $25. What is the sample space for the different amounts of money? Assign probabilities to the sample points and then find the probability that the first envelope purchased contains less than $100.

3. The sample space for the different amounts of money is:

S = {$10, $25, $100}

The probability of each outcome in the sample is calculated as follows:

( ) 55.02011

500275

7515027527510$ ===

++=P ; ( ) 30.0

103

50015025$ ===P ;

( ) 15.0203

50075100$ ===P

The probability that the first envelope purchased contains less than $100 is:

P($10) + P($25) = 0.55 + 0.30 = 0.85

S

8. An automobile manufacturer is concerned about a possible recall of their best-selling four-door sedan. If there were a recall, there is a 0.25 probability that a defect is in the brake system, 0.18 in the transmission system, 0.17 in the fuel system, and 0.40 in some other area.

(a) What is the probability that the defect is in the brakes or the fueling system if the probability of defects in both systems simultaneously is 0.2?

(b) What is the probability that there are no defects in either the brakes or the fueling system?

P

8. Define the following events:

B = Event that the defect is in the brake system; P(B) = 0.25 T = Event that the defect is in the transmission system; P(T) = 0.18 F = Event that the defect is in the fuel system; P(F) = 0.17 O = Event that the defect is in some other area; P(O) = 0.40

(a) The probability that the defect is in the brake or fueling system is:

( ) ( ) ( ) ( ) 22.020.017.025.0 =−+=∩−+=∪ FBPFPBPFBP

Note that P(B∩ F) is actually greater than P(F)-an unusual situation; but the numerical answers are not affected by this.

(b) The probability that there are no defects in either the brake or fueling system is given by:

( ) ( ) ( ) 78.022.011 =−=∪−=∪=∩ FBPFBPFBP

S

15. In a high school graduating class of 100 students, 54 studied mathematics, 69 studied history, and 35 studied both mathematics and history. If one of these students is selected at random, find the probability that

(a) the student took mathematics or history;

P

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(b) the student did not take either of these subjects; (c) the student took history but did not take mathematics.

15. This problem can be visualized using Venn diagram where:

M = Event that student studied mathematics; and H = Event that student studied history.

MS

H

34 1935

12

(a) Probability that the student took mathematics or history is:

( ) ( ) ( ) ( ) 88.010035

10069

10054

=−+=∩−+=∪ HMPHPMPHMP

(b) Probability that the student did not take either of these subjects is:

( ) ( ) 12.088.011 =−=∪−=∪ HMPHMP

(c) Probability that the student took history but did not take mathematics is:

( ) 34.010034

1003569

100)(#)(#

==−

=∩−

=∩HMHHMP

S