Accounting homework help service

Accounting Homework Help Service

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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 accounting problems. Each question demands a different

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

problems our Accounting Homework Help Service 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 represents the event of an unbalanced coin.

From the information given, the prior probabilities P(A1) = .8, and P(A2) = .2 and the

conditional probabilities are :

P(B/A1) = .5 and P(B/A2) = .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 5th 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%.

Question-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 the 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:


Bayesian Probability Table

Events (1) Prior Probabilities

Conditional Probabilities

(3)

Joint Probabilities

Col. 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 number 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 students passing

= 600 + 900 + 60 = 1560

Thus, the probability that the successful student was

An Arts student or P(A1) = 600

1560 = 0385 or 38.5%

a Science student or P(A2) = 900

1560 = .577 or 57.77%

And a Commerce student or P(A3) = 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 :


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

the like chances. The probability of P’s success in the different chances is p(p) = 1

2 +

1

2 ×

1

2 ×

1

2 + ….

= 1

2 [1 +

1

2 2 +

1

2 3 + ….]


Similarly, the probability of Q’s success in his different chances is

P(Q) = 1

2 ×

1

2 +

1

2 ×

1

2 ×

1

2 ×

1

2 +… =

1

4 { 1 +

1

2 2 +

1

2 3+ …. }

Now, the ration of P’s and Q’s probabilities is, 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, the compound probability of A’s success in the 3rd trial has been calculated as under :

Prob. Of A’s failure in the 1st chance × Prob. B’s failure in the 2nd chances × Prob. Of A’s

failure in the 3rd chance = 1

2 ×

1

2 ×

1

2

Similarly, the compound probabilities of B’s success in the 2nd and 4th chances have been as

follows:

2nd chance:

Prob. Of A’s failure in the 1st chance × Prob, of B’s success in the 2nd chance = 1

2 ×

1

2

3rd chance:

Prob. Of A’s failure in the 1st chance × Prob. Of B’s failure in the 2nd Chance × Prob. Of A’s

failure in the 3rd chance × Prob. Of B’s success in the 4th chance = 1

2 ×

1

2 ×

1

2 ×

1

2.


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.


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 p1 , p2, p3, p4 and p5,

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


Thus, the total of his expectations, or

𝑃(ME) = 10

6 +

20

36 +

30

216 +

40

1296 +

50

7776

= 12960 +4320+1080+240+50

7776

= 18650

7776 = $2.40 approx.

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