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  • Accounting Homework Help Service

    Tutorhelpdesk David Luke

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    About Accounting Homework Help Service :

    It has been observed that lack of study technique has been a

    primary reason for downfall of grades in accounting. Some students

    have no choice other than to take accounting as a subject in which

    they have no interest. As a result they end up with a frustrated

    mind before the exams and at the time of submission of their

    accounting assignments. Classroom is not enough to understand

    diverse a ccounting problems. Each question demands a different

    approach to solve. In order to fight against such complex accounts

    problems our Accounting Homework Help S ervice is at your

    service. With our well researched and streamlined study approach we make accounting

    interesting in the mind of students. We follow a step by step method of solving and

    presenting the solution in the simplest manner possible.

    Sample Accounting Assignment Help Service Questions and Answer :

    Depreciation Sample Question

    Question - 1. The probability that a coin is balanced, is .80 and that it is unbalanced, is .20.

    If the coin is balanced, the probability of getting a head is .5, and if the coin is unbalanced

    the probability of a head is .10. The coin is balanced, and that it is unbalanced.

    Solution :

    It is a case of finding the posterior probabilities according to the Bayesian theorem where,

    A1 represents the event of a balanced coin, and

    A2 represen ts the event of an unbalanced coin.

    From the information given, the prior probabilities P(A 1) = .8, and P(A 2) = .2 and the

    conditional probabilities are :

    P(B/A 1) = .5 and P(B/A 2) = .1

    Now, the required posterior probabilities will be computed in a table as under :

    Computation of Posterior Probabilities

    Events (1)

    Prior Probabilities(2)

    Conditional Probabilities (3)

    Joint Probabilities Col.(2 3) (4)

    Posterior Probabilities Col. (4) P (B) (5)

    A1 .8 .5 .40 .4 .42 = .95

    A2 .2 .1 .02 .02 42 = .05

    Total 1.00 - P(B) .42 Total = 1.00

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    From the 5 th column of the above table, it is clear that the probability that the coin is

    balanced is .95 or 95%, and that the coin is unbalanced is .05 or 5%.

    Que stion - 2. A college has been three faculties viz. ,: Arts, Science and Commerce in which

    40% of the students belong to Arts, 50% to Science, and 10% to Commerce. From the

    result of 2003, it was observed that 50% of the Arts students, .60% of the science

    students, and 20% of the commerce students passed in th e examination. If a successful

    students passed in the examination. If a successful student is noticed, what is the

    probability that he was a student of Arts, Science or Commerce?

    Solution :

    Let A1 represent the event of being a student of Arts

    A2 represent the event of being a student of Science

    And A3 represent the event of being a student of Science

    From the information given:

    The prior probabilities are

    A1 = 40% or .4

    A2 = 50% or .5

    A3 = 10% or .1

    The conditional probabilities are:

    P(B/A1 ) = 50%, or .5

    P(B/A2) = 60% , or .6 and

    P(B/A3) = 20%, or .2

    Now, the required posterior probabilities will be calculated in a table as under:

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    Bayesian Probability Table

    Events (1) Prior Probabilities

    Conditional Probabilities

    (3)

    Joint Probabilities

    Co l. 2 3 (4)

    Posterior Probabilities

    Col. 4 P (B) (5)

    (Arts) A1 .4 .5 .1

    .5

    .6

    .2

    .20

    .30

    .02

    .2 .52 = .385 .3 .52 = .577 02 .52 = .038

    Total 1.00 - P(B) = .52 Total = 1.000

    From the 5th columns of the above table, it comes out that the probability

    (i) That the student is an Arts student is .385 or 38.5%

    (ii) That the student is an Science student is .385%

    (iii) That the student is a Commerce student is .038 or 3.8%

    Alternatively:

    Let the total number of students in the college be 3000;

    Then, the num ber of Arts students = 40% of 3000 = 1200;

    The number of Science students = 50% or 3000 = 1500;

    The number of students passing

    Arts = 50% of 1200 = 600;

    Science = 60% of 1500 = 900;

    Commerce = 20% of 300 = 60;

    And therefore, the total number of studen ts passing

    = 600 + 900 + 60 = 1560

    Thus, the probability that the successful student was

    An Arts student or P(A 1) = 600

    1560 = 0385 or 38.5%

    a Science student or P(A 2) = 900

    1560 = .577 or 57.77%

    And a Commerce student or P(A 3) = 60

    1560 = .038 or 3.8%

    (ix) Probability of Mathematical Expectation. Probability of a success multiplied by the

    amount of money one is to get in the event of his success is called the probability of

    Mathematical Expectation in the original sense of the and its related frequency. This may be

    represented as follows :

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    P(ME) = P M, or P F

    Where, p = Probability of success.

    M = Amount of money one is to receive in the event of a success.

    F = Frequency of the event.

    This type of probability is very much used in the games of chance, where efforts are made

    to evaluate the expectation of the players in winning a prize.

    Question - 3 . P and Q throw a coin for a stake of $30 to be won by him, who first flips a

    head. If p is given the first chance, find their respective expectations.

    Solutions

    Here, the respective probabilities of success of P and Q will be determined first as under:

    The probability of flipping a head with a balance coin = 1

    2.

    Here, A can win the 1st, 3rd, 5th and the like chances, and B can win the 2nd, 4th, 6th and 1

    2 +

    1

    2

    1

    2

    1

    2

    = 1

    2 [1 +

    1

    2 2 +

    1

    2

    3

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    P(Q) = 1

    2

    1

    2 +

    1

    2

    1

    2

    1

    2

    1

    2

    1

    4 { 1 +

    1

    2

    2 + 1

    2

    3

    1

    2 :

    1

    4 ; or 2 : 1 i.e.

    2

    3 and

    1

    3.

    Thus their mathematical expectation of the prizes are,

    For P P(ME) = 2

    3 $30 = $20

    For Q P(ME) = 1

    3 $30 = $10

    Note:

    Here,

    failure in the 3rd chance = 1

    2

    1

    2

    1

    2

    Similarly, the compound nd and 4 th chances have been as

    follows:

    2 nd chance:

    st chance nd chance = 1

    2

    1

    2

    3 rd chance:

    st nd

    failure in the 3 rd th chance = 1

    2

    1

    2

    1

    2

    1

    2.

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    Question - 4. In a horse race, a player earns $50,000, if his horse is sound, and $20,000, if

    it is semi -sound. If the probability of the horse being sound is .75, find the expectation of

    the player in the race.

    Solution :

    It is case of mathematical expectation which relates to two mutually exclusive events, i.e.

    either the horse is sound, or the horse is semi -sound.

    If the horse is sound.

    P(A) = .75 $50,000 = $37,500

    If the horse is semi -sound;

    P(B) = .25 $20,000 = $5,000

    P(A B) = P(A) + P(B) = $5,000

    = $42,500

    Hence , the expectation of the player in the race is $42500.

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    Question - 5 . A man throws 5 dice one after another. He will get $10, $20, $30, $40 and

    $50 respectively if he throws 6 each time. He will cease to get anything, when he throws

    any other number. Find the value of his expectation.

    Solution

    Let the probabilities of his success in the 5 trials be represented by p 1 , p 2, p 3, p 4 and p 5,

    and the expectations by p(ME) 1, p(ME) 2 , p(ME) 3, p(ME) 4, and p(ME) 5 respectively.

    Since, the probability of getting a 6 with a balanced die is 1/6,

    P1 = 1/6 = P(ME) 1 = 1/6 $10 = 10

    6

    P2 = 1

    6

    1

    6 = P(ME) 2 =

    1

    36 $20 =

    20

    36

    P3 = 1

    6

    1

    6

    1

    6 = P(ME)3 =

    1

    216 $30 =

    30

    216

    P4 = 1

    6

    1

    6

    1

    6

    1

    6 = P(ME) 4 =

    1

    1296 $40 =

    40

    1296

    And P5 = 1

    6

    1

    6

    1

    6

    1

    6

    1

    6 = P (ME) 5 =

    1

    7776 $50 =

    50

    7776