# Accounting Homework Help Service

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Accounting Homework Help Service

Tutorhelpdesk
David Luke

Phone: (617) 807 0926
Web: http://www.tutorhelpdesk.com
Email: [email protected]
Blog: http://tutorhelpdesk.blogspot.com

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(A 1) = .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)

A1

.8

.5

Joint
Probabilities
Col.(2 × 3)
(4)
.40

A2

.2

.1

.02

Total

1.00

-

P(B) .42

Posterior
Probabilities
Col. (4) ÷ P
(B) (5)
.4 ÷ .42 =
.95
.02 ÷ 42 =
.05
Total = 1.00

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)

(Arts) A1

.4
.5
.1
1.00

.5
.6
.2
-

.20
.30
.02
P(B) = .52

Total

Posterior
Probabilities
Col. 4 ÷ P (B)
(5)
.2 ÷ .52 = .385
.3 ÷ .52 = .577
02 ÷ .52 = .038
Total = 1.000

From the 5th columns of the above table, it comes out that the probability
(i)
(ii)
(iii)

That the student is an Arts student is .385 or 38.5%
That the student is an Science student is .385%
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

a Science student or P(A2) =

= 0385 or 38.5%

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
1
1
the like chances. The probability of P’s success in the different chances is p(p) = +
×
2
2
1
1
2
=

×
1
2

2

+ ….

[1 +

1
2

2

+

1 3
2

+ ….]

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

Now, the ration of P’s and Q’s probabilities is,

1
2

{1+
:

1
4

1 2
2

+

; or 2 : 1 i.e.

1 3
+
2
2
3

and

…. }
1
3

.

Thus their mathematical expectation of the prizes are,
For P → P(ME) =
For Q → P(ME) =

2
3
1
3

× \$30 = \$20
× \$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
1
1
1
failure in the 3rd chance =
× ×
2
2
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
1
1
1
1
failure in the 3rd chance × Prob. Of B’s success in the 4th chance = × ×
× .
2
2
2
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 p 1 , 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 =
P2 =
P3 =
P4 =

1
6
1
6
1
6

×

1
6

×

1

×

1

And P5 =

6
6
1
6

= P(ME)2 =

×

1

×

1

×

1
6

36

×

1

×

1

1
216

20
36

× \$30 =

= P(ME)4 =

6
6

6

× \$20 =

= P(ME)3 =

6
6

1

10

×

1
6

×

1
6

30
216
1
1296

= P (ME)5 =

1
7776

× \$40 =

40
1296

× \$50 =

50
7776

Thus, the total of his expectations, or
𝑃(ME) =

10
6

+

20
36

+

30
216

+

40
1296

=

12960 +4320+1080+240+50

=

18650

+

50
7776

7776
7776

= \$2.40 approx.

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