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    CH02 3

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    yi Lee
    The document contains examples and solutions to probability problems involving mutually exclusive events, conditional probabilities, and Venn diagrams. It analyzes scenarios such as the probabilities of different outcomes for strands of copper wire, skin samples, air conditioning failures, defective orange juice containers, and hospital emergency room visits. For each problem, it clearly shows the probability calculations and logic to arrive at the solutions.

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    © All Rights Reserved
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    CH02 3

    Uploaded by

    yi Lee
    12 views4 pages
    The document contains examples and solutions to probability problems involving mutually exclusive events, conditional probabilities, and Venn diagrams. It analyzes scenarios such as the probabilities of different outcomes for strands of copper wire, skin samples, air conditioning failures, defective orange juice containers, and hospital emergency room visits. For each problem, it clearly shows the probability calculations and logic to arrive at the solutions.

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    Original Title

    CH02-3 (1)

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    The document contains examples and solutions to probability problems involving mutually exclusive events, conditional probabilities, and Venn diagrams. It analyzes scenarios such as the probabilities of different outcomes for strands of copper wire, skin samples, air conditioning failures, defective orange juice containers, and hospital emergency room visits. For each problem, it clearly shows the probability calculations and logic to arrive at the solutions.

    Copyright:

    © All Rights Reserved
    Download as DOCX, PDF, TXT or read online from Scribd
    Download as docx, pdf, or txt
    12 views4 pages

    CH02 3

    Uploaded by

    yi Lee
    The document contains examples and solutions to probability problems involving mutually exclusive events, conditional probabilities, and Venn diagrams. It analyzes scenarios such as the probabilities of different outcomes for strands of copper wire, skin samples, air conditioning failures, defective orange juice containers, and hospital emergency room visits. For each problem, it clearly shows the probability calculations and logic to arrive at the solutions.

    Copyright:

    © All Rights Reserved
    Download as DOCX, PDF, TXT or read online from Scribd
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    of 4

    2.4.1.

    If A, B, and C are mutually exclusive events with P(A) 02, P(B) 03, and P(C) 04, determine the
    following probabilities:

    (a) P(A B C) (b) P(A B  C) (c) P(A B) (d) P[(AB) C] (e) P(ABC)

    (a) P( A∪B∪C ) = P(A) + P(B) + P(C), because the events are mutually exclusive. Therefore,

    P( A∪B∪C ) = 0.2+0.3+0.4 = 0.9

    (b) P ( A∩B∩C ) = 0, because =

    (c) P( A∩B ) = 0 , because =

    (d) P( ( A∪B )∩C ) = 0, because ( A∪B )∩C = ( A∩C )∪( B∩C )=∅

    (e) P( ) =1-[ P(A) + P(B) + P(C)] = 1-(0.2+0.3+0.4) = 0.1

    2.4.6. Strands of copper wire from a manufacturer are analyzed for strength and conductivity. The results from 100

    strands are as follows:

    (a) If a strand is randomly selected, what is the probability that its conductivity is high and its strength is high?

    (b) If a strand is randomly selected, what is the probability that its conductivity is low or its strength is low?

    (c) Consider the event that a strand has low conductivity and the event that the strand has low strength. Are these

    two events mutually exclusive?

    (a) P(High temperature and high conductivity)= 74/100 =0.74

    (b) P(Low temperature or low conductivity)

    = P(Low temperature) + P(Low conductivity) – P(Low temperature and low conductivity)

    = (8+3)/100 + (15+3)/100 – 3/100


    = 0.26

    (c) No, they are not mutually exclusive. Because P(Low temperature) + P(Low conductivity)

    = (8+3)/100 + (15+3)/100

    = 0.29, which is not equal to P(Low temperature or low conductivity).

    2.4.9. A computer system uses passwords that contain exactly eight characters, and each character is one of the 26

    lowercase letters (a–z) or 26 uppercase letters (A–Z) or 10 integers (0–9). Assume all passwords are equally likely.

    Let A and B denote the events that consist of passwords with only letters or only integers, respectively. Determine

    the following probabilities:

    (a) P(A  B) (b) P(A  B) (c) P (Password contains exactly 1 or 2 integers)


    8 8
    52 10
    (a) P(AUB) = P(A) + P(B) =
    8
    + 8 = 0.245
    62 62
    (b) P(A'UB) = P(A') = 1 – 0.2448 = 0.755

    (c) P(contains exactly 1 integer)


    Number of positions for the integer is 8!/(1!7!) = 8

    Number of values for the integer = 10

    Number of permutations of the seven letters is 527

    Total number of permutations is 628


    7
    8 ( 10 ) (52 )
    Therefore, the probability is =¿ 0.377
    628

    P(contains exactly 2 integers)

    Number of positions for the integers is 8!/(2!6!) = 28

    Number of permutations of the two integers is 100

    Number of permutations of the 6 letters is 526

    Total number of permutations is 628


    6
    28 (100 ) (52 )
    Therefore, the probability is =¿ 0.254
    628
    Therefore, P(exactly one integer or exactly two integers) = 0.377 + 0.254 = 0.630

    2.5.2. Samples of skin experiencing desquamation are analyzed for both moisture and melanin content. The results

    from 100 skin samples are as follows:

    Let A denote the event that a sample has low melanin content, and let B denote the event that a sample has high

    moisture content. Determine the following probabilities:

    (a) PA  (b) PB (c) PA | B (d) PB | A


    7+32
    P( A )= =0 . 39
    (a) 100
    13+7
    P( B)= =0 .2
    (b) 100
    P( A∩B ) 7/100
    P( A|B)= = =0 . 35
    (c) P( B) 20/100
    P( A∩B ) 7 /100
    P( B|A )= = =0 .1795
    (d) P( A ) 39/100
    2.5.4. A maintenance firm has gathered the following information regarding the failure mechanisms for air

    conditioning systems:
    The units without evidence of gas leaks or electrical failure showed other types of failure. If this is a representative

    sample of AC failure, find the probability

    (a) That failure involves a gas leak

    (b) That there is evidence of electrical failure given that there was a gas leak

    (c) That there is evidence of a gas leak given that there is evidence of electrical failure

    (a) P(gas leak) = (55 + 32)/107 = 0.813

    (b) P(electric failure | gas leak) = (55/107)/(87/102) = 0.632

    (c) P(gas leak | electric failure) = (55/107)/(72/107) = 0.764

    2.5.6. A batch of 500 containers for frozen orange juice contains 5 that are defective. Two are selected, at random,

    without replacement from the batch.

    (a) What is the probability that the second one selected is defective given that the first one was defective?

    (b) What is the probability that both are defective?

    (c) What is the probability that both are acceptable?

    Three containers are selected, at random, without replacement, from the batch.

    (d) What is the probability that the third one selected is defective given that the first and second ones selected were

    defective?

    (e) What is the probability that the third one selected is defective given that the first one selected was defective and

    the second one selected was okay?

    (f) What is the probability that all three are defective?

    (a) 4/499 = 0.0080


    (b) (5/500)(4/499) = 0.000080

    (c) (495/500)(494/499) = 0.98

    (d) 3/498 = 0.0060

    (e) 4/498 = 0.0080

    (f)

    2.5.7. Suppose A and B are mutually exclusive events. Construct a Venn diagram that contains the three events A,

    B, and C such that PA |C1 and PB |C0.


    2.5.9. Consider the hospital emergency room data given below. Let A denote the event that a visit is to hospital 4,

    and let B denote the event that a visit results in LWBS (at any hospital).

    Determine the following probabilities.

    (a) PA | B (b) PAB (c) PA | B (d) PB | A
    P( A∩B ) 242/22252 242
    P( A|B)= = = =0 . 2539
    (a) P( B) 953 /22252 953
    P( A '∩B) (195+270+246)/22252 711
    P( A '|B)= = = =0 . 7461
    (b) P( B) 953 /22252 953
    P( A∩B ' ) (984 +3103)/22252 4087
    P( A|B' )= = = =0 .1919
    (c) P( B' ) (22252−953)/22252 21299
    P( A∩B ) 242/22252 242
    P( B|A )= = = =0 . 0559
    (d) P( A ) 4329 /22252 4329

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