Types of Office Personnel

An office requires different types of office personnel having different qualifications, skills and knowledge to perform various activities. This note provides you with the brief introduction to office Chief, its functions and about sectional clerk and its functions.

Summary

An office requires different types of office personnel having different qualifications, skills and knowledge to perform various activities. This note provides you with the brief introduction to office Chief, its functions and about sectional clerk and its functions.

Things to Remember

  • An office requires different types of office personnel having different qualifications, skills and knowledge to perform various activities.
  • On the basis of position, responsibilities and nature of the job, office personnel can be classified as follows: Office chief and Sectional chief.
  • The office chief is the in charge who sets the objectives, formulates plans and policies, manages resources, coordinates and controls the whole activities for achieving organizational objectives. 
  • Each department or section is monitored and led by an executive officer, who is known as a departmental or sectional chief. 

MCQs

No MCQs found.

Subjective Questions

Q1:

A basket contains 3 red, 4 black and 5 white balls. A ball is drawn randomly from the basket; find the probability of not getting a black ball.


Type: Short Difficulty: Easy

Show/Hide Answer
Answer: <p>Total balls = 3 red + 4 black + 5 white = 12 balls</p> <p>Let B be the event getting a black ball.</p> <p>Then,</p> <p>P(B) = \(\frac {\text{number of black balls}}{\text{total balls}}\) = \(\frac 4{12}\) = \(\frac 34\)</p> <p>\(\therefore\) P(\(\overline{B}\)) = 1 &ndash; P(B) = 1 - \(\frac 34\) = \(\frac 14\)<sub>Ans</sub></p>

Q2:

A basket contains 5 yellow, 3 blue and 2 green balls. If a ball is drawn randomly from the basket, find the probability of not getting a blue ball.


Type: Short Difficulty: Easy

Show/Hide Answer
Answer: <p>Total no. of balls = 5 + 3 + 2 = 10</p> <p>No. of balls other than blue balls = 5 + 2 = 7</p> <p>Now,</p> <p>P(not getting a blue ball) = \(\frac 7{10}\)<sub>Ans</sub></p>

Q3:

A natural number is chosen at random amongst the first 100. What is the probability that number so chosen is divisible by 3?


Type: Short Difficulty: Easy

Show/Hide Answer
Answer: <p>Here,</p> <p>n = Number of possible cases = 100</p> <p>m = Number of favorable cases = Number of integers from 1 to 100 divisible by 3 = 33</p> <p>\(\therefore\) P(multiple of 3) = \(\frac mn\) = \(\frac {33}{100}\)<sub>Ans</sub></p>

Q4:

From the cards numbered from 5 to 20, a card is drawn at random. Find the probability of getting a card of prime number.


Type: Short Difficulty: Easy

Show/Hide Answer
Answer: <p>The prime numbers from 5 to 20 are 5, 7, 11, 13, 17, 19.</p> <p>\(\therefore\) number of cards of prime numbers = 6</p> <p>Total number of cards numbered from 5 to 20 = 16</p> <p>Now,</p> <p>The probability of getting a card of prime number,</p> <p>P(a prime number) = \(\frac {\text{Number of cards of prime number}}{\text{Total number of cards}}\) = \(\frac 6{16}\) = \(\frac 38\)<sub>Ans</sub></p>

Q5:

A card is drawn from a pack of cards at random. Find the probability of getting a red face card.


Type: Short Difficulty: Easy

Show/Hide Answer
Answer: <p>Here,</p> <p>Number of red face cards = 6</p> <p>Total no. of cards = 52</p> <p>Now,</p> <p>P(a red face card) = \(\frac {\text{Number of red face cards in a pack}}{\text{Total number of cards in a pack}}\) = \(\frac 6{52}\) = \(\frac 3{26}\)<sub>Ans</sub></p>

Q6:

A dice is thrown, find the probability that the face turned up may be even number only.


Type: Short Difficulty: Easy

Show/Hide Answer
Answer: <p>The probability that the face turned may be even number only,</p> <p>P(even number) = \(\frac {\text{the number of even numbered faces in a dice}}{\text{total faces in the dice}}\) = \(\frac 36\) = \(\frac 12\)<sub>Ans</sub></p>

Q7:

Find the probability of not getting 6 when a dice is rolled once.


Type: Short Difficulty: Easy

Show/Hide Answer
Answer: <p>Here,</p> <p>P(getting 6) = \(\frac {\text{number of faces numbered 6}}{\text{total number of faces}}\) = \(\frac 16\)</p> <p>\(\therefore\) P(not getting 6) = 1 &ndash; P(getting 6) = 1 - \(\frac 16\) = \(\frac 56\)<sub>Ans</sub></p>

Q8:

If a cubical die numbered 1 – 6 is thrown, what is the probability that it will show numbers less than 4.


Type: Short Difficulty: Easy

Show/Hide Answer
Answer: <p>The probability that it will show number less than 4 = P(1) + P(2) + P(3) = \(\frac 16\) + \(\frac 16\) + \(\frac 16\) = \(\frac 36\) = \(\frac 12\)<sub>Ans</sub></p>

Q9:

A dice is thrown 1200 times and the record of outcomes is given in the table:

Outcomes

1

2

3

4

5

6

Frequency

186

205

211

187

204

207

Calculate the empirical probability that

  •  less than 4 and
  •  greater than 3.

Type: Short Difficulty: Easy

Show/Hide Answer
Answer: <p>Here,</p> <p>Total number of experiments = sum of the frequencies = 186 + 205 + 211 + 187 + 204 + 207 = 1200</p> <ul><li>Number of outcomes which are less than 4 = 186 + 205 + 211 = 602<br>\(\therefore\) P(number less than 4) = \(\frac {\text{Total number of outcomes}}{\text{Total number of experiments}}\) = \(\frac {602}{1200}\) = \(\frac {301}{600}\)<sub>Ans</sub></li> </ul><ul><li>Number of outcomes greater than 3 = 187 + 204 + 207 = 598<br> \(\therefore\) P(number greater than 3) = \(\frac {\text{number of outcomes}}{\text{total no. of experiments}}\) = \(\frac {598}{1200}\) = \(\frac {299}{600}\)<sub>Ans</sub></li> </ul>

Q10:

From the set of cards numbered 1 to 30, one card is drawn randomly. Find the probability of getting a card divisible by 5 or 7.


Type: Short Difficulty: Easy

Show/Hide Answer
Answer: <p>The cards numbered 5, 10, 15, 20, 25, 30 are divisible by 5, and cards numbered 7, 14, 21, and 28 are divisible by 7.</p> <p>Here,</p> <p>P(divisible by 5) = \(\frac {number\; of\; cards\; that\; are\; divisible\; by\; 5}{total\; number\; of\; cards}\) = \(\frac 6{30}\) = \(\frac 15\)</p> <p>P(divisible by 7) = \(\frac {number\;of\;cards\;that\;are\;divisible\;by\;7}{total\;number\;of\;cards}\) = \(\frac 4{30}\) = \(\frac {2}{15}\)</p> <p>Now,</p> <p>P(divisible by 5 or 7) = P(divisible by 5) + P(divisible by 7) = \(\frac 15\) + \(\frac 2{15}\) = \(\frac {3 + 2}{15}\) = \(\frac {5}{15}\) = \(\frac 13\)<sub>Ans</sub></p>

Q11:

From the number cards numbered from 1 to 30, a card is drawn randomly. Find the probability of getting a card whose number is exactly divisible by 3 or 7.


Type: Short Difficulty: Easy

Show/Hide Answer
Answer: <p>The numbers from 1 to 30 which are exactly divisible by 3 are:</p> <p>3, 6, 9, 12, 15, 18, 21, 24, 27, 30</p> <p>The numbers from 1 to 30 which are exactly divisible by 7 are:</p> <p>7, 14, 21, 28</p> <p>The numbers exactly divisible by both 3 and 7 is only 21.</p> <p>Here,</p> <p>n(S) = 30</p> <p>n(divisible by 3) = 10</p> <p>n(divisible by 7) = 4</p> <p>n(divisible by 3 and 7) = 1</p> <p>Now,</p> <p>\begin{align*} P(divisible\;by\;3\;or\;7) &amp;= P(divisible\;by\;3) + P(divisible\;by\;7) - P(divisible\;by\;3\;and\;7)\\ &amp;= \frac{n(divisible\;by\;3)}{n(S)} + \frac{n(divisible\;by\;7)}{n(S)} - \frac{n(divisible\;by\;3\;and\;7)}{n(S)}\\ &amp;= \frac{10}{30} + \frac {4}{30} - \frac {1}{30}\\ &amp;= \frac {13}{30}_{Ans}\\ \end{align*}</p>

Q12:

From a set of number cards numbered from 1 to 20, a card is drawn randomly. Find the probability of getting a card exactly divisible by 3 or 4.


Type: Short Difficulty: Easy

Show/Hide Answer
Answer: <p>The set of number from 1 to 20 which are exactly divisible by 3 is:</p> <p>A = {3, 6, 9, 12, 15, 18}</p> <p>The set of number from 1 to 20 which are exactly divisible by 4 is:</p> <p>B = {4, 8, 12, 16, 20}</p> <p>From 1 to 20, 12 is only the number exactly divisible by both 3 and 4.</p> <p>i.e. A&cap; B = 12</p> <p>Now,</p> <p>n(S) = 20</p> <p>n(A) = 6</p> <p>n(B) = 5</p> <p>n(A&cap;B) = 1</p> <p>\begin{align*} \therefore P(A\;or\;B) &amp;= P(A) + P(B) - P(A&cap;B)\\ &amp;= \frac {n(A)}{n(S)} + \frac {n(B)}{n(S)} - \frac {n(A&cap;B)}{n(S)}\\ &amp;= \frac 6{20} + \frac 5{20} - \frac 1{20}\\ &amp;= \frac {10}{20}\\ &amp;= \frac 12_{Ans}\\ \end{align*}</p>

Q13:

One card is drawn at random from the number cards, numbered from the number cards, numbered from 5 to 15. Find the probability that card may be prime number or even numbered card.


Type: Short Difficulty: Easy

Show/Hide Answer
Answer: <p>Total numbers from 5 to 15 = 11</p> <p>The prime numbers from 5 to 15 are:</p> <p>5, 7, 11, 13.</p> <p>\(\therefore\) P(getting prime) = \(\frac {prime\;numbers\;from\;5\;to\;15}{numbers\;from\;5\;to\;15}\) = \(\frac 4{11}\)</p> <p>Even numbers from 5 to 15 are:</p> <p>6, 8, 10, 12, 14</p> <p>\(\therefore\) P(getting even) = \(\frac {even\; numbers\;between\;5\;to\;15}{numbers\;from\;5\;to\;15}\) = \(\frac 5{11}\)</p> <p>Here, the events are mutually execlusive.</p> <p>Hence,</p> <p>P(getting prime or even numbers) = P(prime numbers) + P(even numbers) = \(\frac 4{11}\) + \(\frac 5{11}\) = \(\frac {9}{11}_{Ans}\)</p>

Q14:

A card is drawn from a well-shuffled pack of 52 cards. Find the probability of getting a queen or a black ace.


Type: Short Difficulty: Easy

Show/Hide Answer
Answer: <p>Here,</p> <p>n(S) = 52</p> <p>n(Queen) = 4</p> <p>n(Black ace) = 2</p> <p>Hence,</p> <p>The probability of getting a queen or black ace is:</p> <p>\begin{align*} P(queen\;or\;black\;ace) &amp;= P(Queen) + P(Black\; ace)\\ &amp;= \frac {n(Queen)}{n(S)} + \frac {n(Black\; ace)}{n(S)} \\ &amp;= \frac 4{52} + \frac 2{52}\\ &amp;= \frac {4 + 2}{52}\\ &amp;= \frac {6}{52}\\ &amp;= \frac {3}{26}_{Ans}\\ \end{align*}</p>

Q15:

From a pack of cards, a card is drawn at random. Find the probability of getting this card a black or non- faced.


Type: Short Difficulty: Easy

Show/Hide Answer
Answer: <p>Suppose, B denotes the event getting black cards and F denotes the event getting face cards.</p> <p>\(\therefore\) P(B) = \(\frac {number\;of\;black\;cards\;in\;a\;pack}{number\;of\;cards\;in\;a\;pack}\) = \(\frac {26}{52}\) = \(\frac 12\)</p> <p>and P(F) =\(\frac {number\;of\;face\;cards\;in\;a\;pack}{number\;of\;cards\;in\;a\;pack}\) = \(\frac {12}{52}\) = \(\frac 3{13}\)</p> <p>Then,</p> <p>The probability getting not a faced card:</p> <p>P(\(\overline {F}\)) = 1 - \(\frac 3{13}\) = \(\frac {10}{13}\)</p> <p>Again,</p> <p>The events B and \(\overline {F}\) are not mutually exclusive events.</p> <p>So;</p> <p>P(B&cap;\(\overline{F}\)) =\(\frac {number\;of\;black\;non\;faced\;cards\;in\;a\;pack}{number\;of\;cards\;in\;a\;pack}\) = \(\frac {20}{52}\) = \(\frac 5{13}\)</p> <p>By the formula,</p> <p>P(B&cup;\(\overline{F}\)) = P(B) + P(\(\overline {F}\)) -P(B&cap;\(\overline{F}\)) = \(\frac 12\) + \(\frac {10}{13}\) - \(\frac 5{13}\) = \(\frac {13+20-10}{26}\) = \(\frac {23}{26}_{Ans}\)</p>

Q16:

A bag contains 4 black and 3 white balls. A ball is drawn randomly and not replaced then another ball is drawn. What is the probability of getting two black balls?


Type: Short Difficulty: Easy

Show/Hide Answer
Answer: <p>Here,</p> <p>Total number of balls = 4 black + 3 white = 7 balls</p> <p>Now,</p> <p>The probability of getting first black ball;</p> <p>P<sub>1</sub>(Black) = \(\frac {number\;of\;black\;balls}{total\;number\;of\;balls}\) = \(\frac 47\)</p> <p>If the first ball is drawn and not replaced, the probability of getting second black ball,</p> <p>P<sub>2</sub> (Black) = \(\frac {number\;of\;remaining\;black\;balls}{total\;number\;of\;remaining\;balls}\) = \(\frac 36\) = \(\frac 12\)</p> <p>Therefore,</p> <p>P(black ball/black ball) = P<sub>1</sub> (black ball)&times; P<sub>2</sub> (black ball) = \(\frac 47\)&times; \(\frac 12\) = \(\frac 27\)<sub>Ans</sub></p>

Q17:

A box contains 6 red, 4 white and 5 blue balls. From this box 3 balls are drawn in succession. Find the probability that they are drawn in the order red, white and blue if each ball is repeated.


Type: Short Difficulty: Easy

Show/Hide Answer
Answer: <p>Total number of balls = 6 + 4 + 5 = 15</p> <p>Probability of getting first ball is red,</p> <p>P<sub>1</sub> (red) = \(\frac 6{15}\) = \(\frac 25\)</p> <p>Probability of getting second ball is white,</p> <p>P<sub>2</sub> (white) = \(\frac 4{15}\), since first ball is replaced.</p> <p>Probability of getting third ball is blue,</p> <p>P<sub>3</sub> (blue) = \(\frac 5{15}\) = \(\frac 13\), since first and second ball is replaced.</p> <p>&there4; P (red, white, blue) = P<sub>1</sub> (red) &times;P<sub>2</sub> (white) &times;P<sub>3</sub> (blue) = \(\frac 25\) &times;\(\frac 4{15}\) &times;\(\frac 13\) = \(\frac 8{255}\)<sub>Ans</sub></p>

Q18:

A card is drawn at random from a pack of 52 cards and at the same time a marble is drawn at random from a bag containing 2 red marbles and 3 blue marbles. Find the probability of getting a blue marble and a king.


Type: Short Difficulty: Easy

Show/Hide Answer
Answer: <p>Here,</p> <p>Number of kings in a pack of cards = 4</p> <p>Total no. of cards in the pack = 52</p> <p>P(a king) = \(\frac {number\;of\;kings\;in\;a\;pack\;of\;cards}{total\;number\;of\;cards\;in\;a\;pack\;of\;cards}\) = \(\frac 4{12}\) = \(\frac 13\)</p> <p>Again,</p> <p>Number of blue marbles = 3</p> <p>Total number of marbles = 2 red + 3 blue = 5</p> <p>P(a blue marble) = \(\frac {number\;of\;blue\;marbles}{total\;number\;of\;marbles}\) = \(\frac 35\)</p> <p>&there4; P (a blue marble and a king) = P(a blue marble) + P(a king) = \(\frac 35\) + \(\frac 1{13}\) = \(\frac {39 + 5}{65}\) = \(\frac {44}{65}\)<sub>Ans</sub></p> <p></p>

Q19:

A card is drawn at random from a well-shuffled pack of 52 cards and again it is put on the same pack and a card is drawn from the pack find the probability that  the two cards are:

  1. both diamond
  2. diamond and heart

Type: Short Difficulty: Easy

Show/Hide Answer
Answer: <p>Here,</p> <p>Total number of cards = 52</p> <ol><li>In the first time, <br>P(a diamond) = \(\frac {13}{52}\) = \(\frac 14\)<br>In the second time, <br>P(a diamond) = \(\frac {13}{52}\) = \(\frac 14\)<br>&there4; P(diamond both times) = P(a diamond)&times; P(a diamond) = \(\frac 14\)&times; \(\frac 14\) = \(\frac 1{16}\)<sub>Ans</sub></li> <li>In the first time, <br>P(a diamond) = \(\frac {\text{number of diamond in the pack}}{\text{total number of cards}}\) = \(\frac {13}{52}\) = \(\frac 14\)<br>In the second time,<br>P(a heart) = \(\frac {\text{number of heart in the pack}}{\text{total number of cards}}\) = \(\frac {13}{52}\) = \(\frac 14\)<br>&there4; P(a diamond and a heart) = P(a diamond)&times; P(a heart) = \(\frac 14\)&times; \(\frac 14\) = \(\frac 1{16}\)<sub>Ans</sub></li> </ol>

Q20:

Find the probability of getting a tail on the coin and 4 on the dice, when a coin is tossed and a dice is rolled simultaneously.


Type: Short Difficulty: Easy

Show/Hide Answer
Answer: <p>When a coin is tossed, the sample space S has two sample points, S = {H, T} where H and T denote Head and Tail respectively.</p> <p>&there4; n(S) = 2</p> <p>Now,</p> <p>P(T) = \(\frac {n(T)}{n(S)}\) = \(\frac 12\)</p> <p>Since, in a single throw of a dice, the sample space S has six sample points,</p> <p>S = {1, 2, 3, 4, 5, 6}</p> <p>&there4; n(S) = 6</p> <p>Here,</p> <p>P(4) = \(\frac {\text{number of face 4 in S}}{n(S)}\) = \(\frac 16\)</p> <p>Therefore,</p> <p>P(T and 4) = P(T)&times; P(4) = \(\frac 12\)&times; \(\frac 16\) = \(\frac 1{12}\)<sub>Ans</sub></p>

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Types of Office Personnel

Types of Office Personnel

An office requires different types of office personnel having different qualifications, skills and knowledge to perform various activities. Some personnel work at the top level, some work at a middle level and other works at a lower level. On the basis of position, responsibilities and nature of job, office personnel can be classified as follows:

Office Chief

The office chief is the in charge who sets the objectives, formulates plans and policies, manages resources, coordinates and controls the whole activities for achieving organizational objectives. He is regarded as the boss of the organization. He is the person who is fully responsible for the successful operation of the organization. He is the person who plays the role of the pilot of plane, captain of a ship and driver of a bus.

Functions of Office Chief

  1. Setting objectives of the organization.
  2. Preparing long term and short terms plans and policies of the organization.
  3. Dividing the work among the workers in the organization.
  4. Directing and coordinating the activities of the different staff and departments.
  5. Adopting good remuneration system for motivating workers.
  6. Maintaining a sound accounting system in the organization.
  7. Reporting the performance and achievement of the organization to the board and public.
  8. Evaluating the performance of the staff and departments.

Needs of Office Chiefs

Following points highlight the necessity and importance of Office Chief:

  1. To determine the objectives of the organization.
  2. To make plans, policies, programs, rules and regulations for achieving the set objectives.
  3. To motivate employees in their works by providing attractive remuneration, facilities and opportunities.
  4. To maintain discipline in the office and work.
  5. To form employee development to increase the efficiency of the staff.
  6. To avoid delay in office procedures through appropriate supervision, direction and institution.
  7. To make the proper evolution of the performance of employees for their compensation and promotion.

Sectional Chief

An organization is divided into different sections or departments on the basis of their functions. These departments are production, finance, personnel, marketing, accounting and public relation departments. Each department or section is monitored and led by an executive officer, who is known as a departmental or sectional chief. He is a middle-level officer of the organization. He is the officer who performs all the activities of the concerned department under the direction of the office chief. He prepares plans and policies, makes a division of work and mobilizes the resources of the department efficiently.

Functions of Sectional Chief

  1. Setting the departmental objectives in conformity with the organizational objectives.
  2. Preparing departmental plans and policies for achieving departmental objectives.
  3. Reporting the performance, problem and achievement of the department to the office chief.
  4. Dividing the work among the staff working in the department.
  5. Mobilizing human and other resources in an efficient manner for carrying out departmental activities smoothly.

Lesson

Office Personnel

Subject

Accountancy

Grade

Grade 9

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