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Edited By Ravindra Pindel | Updated on Sep 15, 2022 05:25 PM IST | #CBSE Class 12th

**NCERT exemplar Class 12 Maths solutions chapter 13** deals with the possibility of occurrence of any event or explains how likely it is for an event to occur. The possibility of occurring or not occurring of any event cannot be predicted but can be displayed in the form of probability with the absolute level of certainty. Probability of any event is a number between 0 to 1 where 0 shows the absolute impossibility of an event, and 1 shows the maximum chances of happening of an event certainly. NCERT exemplar Class 12 Maths solutions chapter 13 help students to interrelate such knowledge with real-life problems and display the application of such knowledge in different lives scenarios and situations along with enhancing decision-making at an efficient level.

**Also, read -** NCERT Class 12 Maths Solutions

Question:1

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Answer:

Given

For a loaded die –

P (1) = P (2) = 0.2, P (3) = P (5) = P (6) = 0.1 and P (4) = 0.3

The die is thrown twice and

- Event of the same number each time
- Event of total score of 10 or more.

Therefore, for A,

For B,

B= EVENT OF TOTAL SCORE IS 10 OR MORE

B= {(4,6), (5,5), (5,6), (6,4), (6,5), (6,6)}

For the probability of intersection B i.e. both the events occur simultaneously,

Therefore,

P (A B) = P (5,5) + P (6,6)

P (A B) = P (5) × P (5) + P (6) ×P (6)

P (A B) = 0.1×0.1+ 0.1×0.1

P (A B) = 0.01+0.01

P (A B) = 0.02

Knowing that if two events are independent, then,

Therefore,

Hence, A and B are independent events.

Question:2

Answer:

Given

Therefore,

And

Therefore,

Therefore,

Therefore,

P (A B) ≠ P(A). P(B)

Hence, A and B are not independent events.

Question:3

Answer:

Given-

At least one of the two events A and B occurs is 0.6 i.e. P(AB) = 0.6

If A and B occur simultaneously, the probability is 0.3 i.e. P(AB) = 0.3

It is known to us that

P(AB) = P(A)+ P(B) – P(AB)

0.6 = P(A)+ P(B) – 0.3

P(A)+ P(B) = 0.6+ 0.3 = 0.9

To find-

Therefore,

Question:4

Answer:

Given

A bag contains 5 red marbles and 3 black marbles

If the first marble is red, the following conditions have to be followed for at least one marble to be black.

- Second marble is black and third is red =
- Second and third, both marbles are black =

(iii) Second marble is red and third marble is black =

Let event = drawing red marble in draw

And event = drawing black marble in draw

Hence, the for the probability for at least one marble to be black is P

Question:5

Answer:

Given-

Two dice are drawn together i.e. n(S)= 36

S is the sample space

E = a of total 4

F= a total of 9 or more

G= a total divisible by 5

Therefore, for E,

E = a of total 4

∴E = {(2,2), (3,1), (1,3)}

∴n(E) = 3

For F,

F= a total of 9 or more

∴ F = {(3,6), (6,3), (4,5), (5,4), (6,4), (4,6), (6,5), (6,6), (5,5), (5,6)}

∴n(F)=10

For G,

G = a total divisible by 5

∴ G = {(1,4), (4,1), (2,3), (3,2), (4,6), (6,4), (5,5)}

∴ n(G) = 7

Here, (E F) = φ AND (E G) = φ

Also, (F G) = {(4,6), (6,4), (5,5)}

n (F G) = 3 and (E F G) = φ

Therefore,

P (F G) ≠ P(F). P(G)

Hence, there is no pair which is independent

Question:6

Explain why the experiment of tossing a coin three times is said to have binomial distribution.

Answer:

Let p= events of failure and q=events of success

It is known to us that,

A random variable X (=0,1, 2,…., n) is said to have Binomial parameters n and p, if its probability distribution is given by

Therefore, in the experiment of a coin being tossed three times,

we have random variable X which can take values 0,1,2 and 3 with parameters n=3 and

Hence, tossing of a coin 3 times is a Binomial distribution.

Question:7

A and B are two events such that

Find:

(i) P(A|B) (ii) P(B|A) (iii) P(A’|B) (iv) P(A’|B’)

Answer:

Given-

iv)

Question:9

By De Morgan's laws

Knowing that

Hence, either

(iv)

Hence, either

Question:11

Answer:

It has to be proven that

As we know,

Therefore,

When two events cannot occur at the same time, they are called mutually exclusive or disjoint events.

Since () means A and B both occurring at the same time and

() means A and both occurring at the same time.

Therefore, it is not possible that occur at the same time.

Hence, are mutually exclusive.

∴ P[(A B) (A B’)] = 0 ….. (1)

Therefore, A = A (B B’)

Knowing that P (A B) = P (A) + P(B) - P (A B)

Hence proved.

ii) It is to be proven that,

AB means the all the possible outcomes of both A and B.

From the above Venn diagram,

When two events cannot occur at the same time, they are called mutually exclusive or disjoint events.

Since (A B) means A and B both occurring at the same time and

(A B’) means A and B’ both occurring at the same time.

(A' B) means A’ and B both occurring at the same time.

Therefore, it is not possible that (A B), (A B’) and ( B) occur at the same time.

Hence these events are mutually exclusive.

It is known that,

Therefore,

From (1),(2),(3) and (4) we get,

Hence Proved

Question:12

If X is the number of tails in three tosses of a coin, determine the standard deviation of X.

Answer:

Given-

Radom variable X is the member of tails in three tosses of a coin

Therefore, X= 0,1,2,3

Substituting the values of in equation (i), we get:

And standard deviation of

Question:13

Answer:

Let X = the random variable of profit per throw

Probability of getting any number on dice is .

Since, she loses Rs 1 on getting any of 2, 4 or 5.

Therefore, at X= -1,

P(X) = P (2) +P(4) +P(5)

In the same way, =1 if dice shows of either 1 or 6 .

and at X=4 if die shows a 3

Question:14

Answer:

Since three dice are thrown at the same time, the sample space is [n(S)] = 6^{3}= 216.

Let E_{1} be the event when the sum of numbers on the dice was six and

E_{2} be the event when three twos occur.

Question:15

Answer:

Let X be the variable for the prize

The possibility is of winning nothing, Rs 500, Rs 2000 and Rs 3000.

So, X will take these values.

Since there are 3 third prizes of 500, the probability of winning third prize is .

1 first prize of 3000, so probability of winning third prize is .

1 second prize of 2000, so probability of winning third prize is .

Question:16

Answer:

Given-

= [4 white balls] and = [5 black balls]

= [9 white balls] and = [7 black balls]

Let be the event that the ball transferred from the first bag is white and

be the event that the ball transferred from the bag is black.

E is the event that the ball drawn from the second bag is white.

Question:17

Answer:

Given-

Bag I= [3Black, 2White], Bag II= [2 black, 4 white]

Let E_{1} be the event that bag I is selected

E_{2} be the event that bag II is selected

E_{3} be the event that a black ball is selected

Therefore,

Question:18

Answer:

Given-

The box has 5 blue and 4 red balls.

Let E_{1} be the event that first ball drawn is blue

E_{2 }be the event that the first ball drawn is red and

E be the event that second ball drawn is blue.

Question:19

Answer:

Let E_{1}, E_{2}, E_{3 }and E_{4} be the events that the first, second, third and fourth card is king respectively.

As we know, there are 4 kings,

when 1 king is taken out, there are 3 kings and total 51 cards left.

Therefore, probability of drawing a king when one king has been taken out is:

When 2 kings are taken out, there are 2 kings and 50 cards left. Therefore, probability of drawing a king when two kings have been taken out is:

When 3 kings are taken out, 1 king and 49 cards are left.

Therefore, probability of drawing a king when three kings have been taken out is:

Probability that all 4 cards are king is:

Question:20

A die is thrown 5 times. Find the probability that an odd number will come up exactly three times.

Answer:

Given-

n=5, Odd numbers = 1,3,5

and

also, r=3

Question:21

Ten coins are tossed. What is the probability of getting at least 8 heads?

Answer:

Let X = the random variable for getting a head.

Here, n=10, r≥8

r=8,9,10

Question:22

Answer:

Given-

The man shoots 7 times, therefore, n=7

And probability of hitting target is

Question:23

Answer:

Given-

There are 10 defective watches in 100 watches

The probability of defective watch from a lot of 100 watch

As we know that

Question:24

Consider the probability distribution of a random variable X:

Calculate (i) (ii) Variance of X.

Answer:

Given-

Question:25

The probability distribution of a random variable X is given below:

(i) Determine the value of k.

(ii) Determine P (X ≤ 2) and P (X > 2)

(iii) Find P (X ≤ 2) + P (X > 2).

Answer:

Given-

(i) since,

(ii)

Question:26

For the following probability distribution determine standard deviation of the random variable X.

Answer:

Given-

Question:27

Answer:

Given-

X= number of four seen

On tossing to die, X=0,1,2

Therefore,

Thus, the table is derived

Question:28

A die is thrown three times. Let X be ‘the number of twos seen’. Find the expectation of X.

Answer:

Given-

X= no. of twos seen

Therefore, on throwing a die three times, we will have X=0,1,2,3

Question:29

Answer:

Given-

For the second die

Let x be the number of ones seen

For X=0,

Question:30

Two probability distributions of the discrete random variable X and Y are given below.

Prove that .

Answer:

To prove that-

Taking LHS of equation (i), we have:

Taking RHS of equation (i) we get:

Thus, from equations (ii) and (iii), we get:

Hence proved.

Question:31

Answer:

Let X be the random variable which denotes that the bulb is defective.

And

(j) None of the bulbs is defective i.e., r=0

(ii)Exactly two bulbs are defective i.e., r=2

Question:32

Answer:

Let E_{1} be the event that a fair coin is drawn

E_{2} be the event that two headed coin is drawn

E be the event that tossed coin get a head

Question:33

Answer:

Given-

Let E_{1} be the event that the person selected is of group O

E_{2} be the event that the person selected is of other than blood group O

And E_{3} be the event that the person selected is left handed

∴P(E_{1}) =0.30, P(E_{2}) =0.70

P(E_{3}|E_{1}) = 0.060 And P(E_{3}|E_{2}) =0.10

Using bayes’ theorem, we have:

Question:34

Answer:

The notation P [r ≤ p| s ≤ p] means that

P (r ≤p) *given that s ≤ p

Since we know s ≤ p , then it means that s is drawn first.

Let us have n numbers before s is drawn:

(1 . . s …. . p . . .. n)

After s is drawn,

[ 1 ... p] has one element missing, so there are (p-1) elements.

Also, there is one element missing from the entire set, so there are (n-1) altogether.

Among (1 . . s …. . p) the probability of drawing s is .

is the probability that

Therefore,

Question:35

Answer:

Let X be the random variable score obtained when a die is thrown twice.

∴ X= 1,2,3,4,5,6

The sample space is

In the same way,

Therefore, the required distribution is,

Question:36

Answer:

Given-

X=0,1,2 and P(X) at X=0 and 1,

Let X=2, P(X) is x.

The following distribution is obtained

And

And, as we know that

Question:38

Answer:

Given-

A and B throw a pair of dice alternately.

A wins if he gets a total of 6

And B wins if she gets a total of 7

Therefore,

A = {(2,4), (1,5), (5,1), (4,2), (3,3)} and

B = {(2,5), (1,6), (6,1), (5,2), (3,4), (4,3)}

Let P(B) be the probability that A wins in a throw

And P(B) be the probability that B wins in a throw

The probability of A winning the game in the third row

Question:39

Answer:

Given-

A= {(x, y):x+y=11}

And B= {(x, y): x≠5}

∴ A = {(5,6), (6,5)}

B= {(1,1), (1,2), (1,3), (1,4), ((1,5), (1,6), (2,1), (2,2), (2,3), (2,4), (2,5), (2,6), (3,1), (3,2), (3,3), (3,4), (3,5), (3,6), (4,1), (4,2), (4,3), (4,4), (4,5), (4,6), (6,1), (6,2), (6,3), (6,4), (6,5), (6,6)}

Question:40

Answer:

Given-

An urn contains m white and n black balls.

Let E_{1} be the first ball drawn of white colour

E_{2} be the first ball drawn of black colour

And E_{3} be the second ball drawn of white colour

Using the probability theorem, we get,

Hence, the probability of drawing a white ball does not depend on k.

Question:41

Answer:

Let E_{1}, E_{2}, and E_{3} be the events that Bag 1, Bag 2 and Bag 3 are selected, and a ball is chosen from it.

Bag 1: 3 red balls,

Bag 2: 2 red balls and 1 white ball

Bag 3: 3 white balls.

The probability that bag i will be chosen and a ball is selected from it is i|6.

The Law of Total Probability:

In a sample space S, let E_{1},E_{2},E_{3}…….E_{n} be n mutually exclusive and exhaustive events associated with a random experiment. If A is any event which occurs with E_{1},E_{2},E_{3}…….E_{n}, then

(i) Let “E” be the event that a red ball is selected.

P(E|E1) is the probability that red ball is chosen from the bag 1.

P(E|E2) is the probability that red ball is chosen from the bag 2.

P(E|E3) is the probability that red ball is chosen from the bag 3.

Therefore,

As red ball can be selected from Bag 1, Bag 2 and Bag 3.

Therefore, probability of choosing a red ball is the sum of individual probabilities of choosing the red from the given bags.

From the law of total probability,

Let F be the event that a white ball is selected.

Therefore, P(F|E1) is the probability that white ball is chosen from the bag 1.

P(F|E2) is the probability that white ball is chosen from the bag 2.

P(F|E3) is the probability that white ball is chosen from the bag 2.

P(F|E1) = 0

As white ball can be selected from Bag 1, Bag 2 and Bag 3

Therefore, sum of individual probabilities of choosing the red from the given bags is the probability of choosing a white ball.

Question:42

Answer:

Referring to the previous question, using Bayes theorem, we get

Let E_{1}, E_{2}, and E_{3} be the events that Bag 1, Bag 2 and Bag 3 is selected, and a ball is chosen from it.

Bag 1: 3 red balls,

Bag 2: 2 red balls and 1 white ball

Bag 3: 3 white balls.

The probability that bag i will be chosen and a ball is selected from it is i|6.

Let F be the event that a white ball is selected. Therefore, is the probability that white ball is chosen from the bag 1 .

is the probability that white ball is chosen from the bag 2 .

To find: the probability that if white ball is selected, it is selected from:

(i) Bag 2

Using Bayes’ theorem to find the probability of occurrence of an event A when event B has already occurred.

(ii)Bag 3

Using Bayes' theorem to find the probability of occurrence of an event A when event B has already occurred.

Using Bayes' theorem, we get the probability of as:

Question:43

Given-

A1, A2, and A3 denote the three types of flower seeds and A1: A2: A3 = 4: 4 : 2

Therefore, Total outcomes = 10

Therefore,

Let E be the event that a seed germinates and E’ be the event thata seed does not germinate.

P(E|A1) is the probability that seed germinates when it is seed A1.

P(E’|A1) is the probability that seed will not germinate when it is seed A1.

P(E|A2) is the probability that seed germinates when it is seed A2.

P(E’|A2) is the probability that seed will not germinate when it is seed A2.

P(E|A3) is the probability that seed germinates when it is seed A3.

P(E’|A3) is the probability that seed will not germinate when it is seed A3.

The Law of Total Probability:

In a sample space mutually exclusive and exhaustive events associated with a random experiment. If A is any event which occurs with then

(i) Probability of a randomly chosen seed to germinate.

It can be either seed A, B or C, therefore,

From law of total probability,

(ii) that it will not germinate given that the seed is of type A3

Knowing that P(A) + P(A’) =1

P(E’|A3) = 1 – P(E|A3)

(iii) that it is of the type A2 given that a randomly chosen seed does not germinate.

Using Bayes’ theorem to find the probability of occurrence of an event A when event B has already occurred.

Question:44

Answer:

Let events E1, E2 be the following events-

E1 be the event that letter is from TATA NAGAR and E2 be the event that letter is from CALCUTTA

Let E be the event that on the letter, two consecutive letters TA are visible.

Since, the letter has come either from CALCUTTA or TATA NAGAR

We get the following set of possible consecutive letters when two consecutive letters are visible in the case of TATA NAGAR

{TA, AT, TA, AN, NA, AG, GA, AR}

We get the following set of possible consecutive letters in the case of CALCUTTA,

{CA, AL, LC, CU, UT, TT, TA}

Therefore, P(E|E1) is the probability that two consecutive letters are visible when letter came from TATA NAGAR

P(E|E2) is the probability that two consecutive letters are visible when letter came from CALCUTTA

To find- the probability that the letter came from TATA NAGAR.

Using Bayes’ theorem to find the probability of occurrence of an event A when event B has already occurred.

Question:45

Answer:

Given-

There are 2 bags-

Bag 1: 3 black and 4 white balls

Bag 2: 4 black and 3 white balls

Therefore, Total balls = 7

Let events E1, E2 be the following:

E1 and E2 be the events that bag 1 and bag 2 are selected respectively

We know that a die is thrown.

Therefore, total outcomes = 6

The Law of Total Probability:

In a sample space S, let E1,E2,E3…….En be n mutually exclusive and exhaustive events associated with a random experiment. If A is any event which occurs with E1,E2,E3…….En, then

P(A) = P(E1)P(A/E1)+ P(E2)P(A/E2)+ …… P(En)P(A/En)

Let “E” be the event that black ball is chosen.

P(E|E1) is the probability that black ball is chosen from the bag 1.

P(E|E2) is the probability that black ball is chosen from the bag 2.

Therefore,

Therefore, probability of choosing a black ball is the sum of individual probabilities of choosing the black from the given bags.

From the law of total probability,

Question:46

Answer:

Given-

There are 3 urns U1, U2 and U3

Let U1 be 2 white and 3 black balls

U2 be 3 white and 2 black balls

U3 be 4 white and 1 black balls

Therefore, Total balls = 5

As there is an equal probability of each urn being chosen

Let be the event that a ball is chosen from an urn

Let A be the event that white ball is drawn.

P(A|E1) is the probability that white ball is chosen from urn U1

P(A|E2) is the probability that white ball is chosen from urn U2

P(A|E3) is the probability that white ball is chosen from urn U3

To find- the probability that the ball is drawn was from

Using Bayes' theorem to find the probability of occurrence of an event A when event B has already occurred.

Question:47

Answer:

Let events E1, E2, E3 be the following events:

E1 - the event that person has TB and E2 - the event that the person does not have TB

Therefore, Total persons = 1000

Therefore,

Let E be the event that the person is diagnosed to have TB

To find- the probability that the person actually has TB

Using Bayes' theorem to find the probability of occurrence of an event A when event B has already occurred.

is the probability that person actually has TB

Question:48

Answer:

Let E1, E2, E3 be the following events:

E1 -event that item is manufactured by machine A

E2- event that item is manufactured by machine B

E3- event that item is manufactured by machine C

Since, E1, E2 and E3 are mutually exclusive and exhaustive events hence, they represent a partition of sample space.

As we know that

Items manufactured on machine A = 50%

Items manufactured on machine B = 30%

Items manufactured on machine C = 20%

Therefore,

Let E be the event that ‘an item is defective’.

Therefore, P(E|E1) is the probability of the item drawn is defective given that it is manufactured on machine A = 2%

P(E|E2) is the probability of the item drawn is defective given that it is manufactured on machine B = 2%

P(E|E3) is the probability of the item drawn is defective given that it is manufactured on machine C = 3%

Therefore,

To find- the probability that the item which is picked up is defective, it was manufactured on machine A

Using Bayes’ theorem to find the probability of occurrence of an event A when event B has already occurred.

is the probability that the item is drawn is defective and it was manufactured on machine A

Question:49

Let X be a discrete random variable whose probability distribution is defined as follows:

where k is a constant. Calculate

(i) the value of k (ii) E (X) (iii) Standard deviation of X.

Answer:

Given-

Therefore, we get the probability distribution of X as

(i) the value of k

As we know, Sum of the probabilities =1

To find: E(X)

The probability distribution of X is:

(iii) To find: Standard deviation of X

As we know,

As we know,

standard deviation of

Question:50

The probability distribution of a discrete random variable X is given as under:

Calculate :

(i) The value of A if E(X) = 2.94

(ii) Variance of X.

Answer:

i ) Given-

E(X) = 2.94

It is known to us that μ = E(X)

x1

Question:51

The probability distribution of a random variable x is given as under:

where k is a constant. Calculate

(i) E(X) (ii) (iii) P(X ≥ 4)

Answer:

Given-

As we know, Sum of the probabilities =1

(i) To find:

or

(ii) To find:

We first find

As we know that,

Question:52

Answer:

Given-

n coins have head on both the sides and (n + 1) coins are fair coins

Therefore, Total coins = 2n + 1

Let E1, E2 be the following events:

E1 = Event that an unfair coin is selected

E2 = Event that a fair coin is selected

The Law of Total Probability:

In a sample space S, let E1,E2,E3…….En be n mutually exclusive and exhaustive events associated with a random experiment. If A is any event which occurs with E1,E2,E3…….En, then

P(A) = P(E1)P(A/E1)+ P(E2)P(A/E2)+ …… P(En)P(A/En)

Let “E” be the event that the toss result is a head

P(E|E1) is the probability of getting a head when unfair coin is tossed

P(E|E2) is the probability of getting a head when fair coin is tossed

Therefore,

Question:53

Answer:

Let X be a random variable of number of aces

X can take values 0, 1 or 2 because only two cards are drawn.

Therefore, Total deck of cards = 52

and total no. of ACE cards in a deck of cards = 4

Since the draws are done without replacement, therefore, the two draws are not independent.

Therefore,

P(X = 0) = Probability of no ace being drawn

= P(non – ace and non – ace)

= P(non – ace) × P(non – ace)

Question:54

Answer:

Let X be the random variable for a ‘success’ for getting an even number on a toss.

∴ X = 0, 1, 2

n = 2

Even number on dice = 2, 4, 6

∴ Total possibility of getting an even number = 3

Total number on dice = 6

p = probability of getting an even number on a toss

Question:55

Answer:

The sample space is

S = { (1,2),(1,3),(1,4),(1,5)

(2,1),(2,3),(2,4),(2,5)

(3,1),(3,2),(3,4),(3,5)

(4,1),(4,2),(4,3),(4,5)

(5,1),(5,2),(5,3),(5,4)}

Total Sample Space, n(S) = 20

Let random variable be X which denotes the sum of the numbers on the cards drawn.

∴ X = 3, 4, 5, 6, 7, 8, 9

At X = 3

The cards whose sum is 3 are (1,2), (2,1)

At x=4

The cards whose sum is 4 are (1,3),(3,1)

At X=5

The cards whose sum is 5 are (1,4),(2,3),(3,2),(4,1)

At X=6

The cards whose sum is 6 are (1,5),(2,4),(4,2),(5,1)

At x=7

The cards whose sum is 7 are (2,5),(3,4),(4,3),(5,2)

At X = 8

The cards whose sum is 8 are (3,5), (5,3)

At X = 9

The cards whose sum is 9 are (4,5), (5,4)

Question:56

If , and , then P(B | A) is equal to

A. 1|10

B. 1|8

C. 7|8

D. 17|20

Answer:

Given- , and

Hence, Correct option is C

Question:57

If P(A ∩ B) = 7|10 and P(B) = 17|20, then P(A|B) equals

A. 14|17

B. 17|20

C. 7|8

D. 1|8

Answer:

Given-

Hence, Correct option is A

Question:58

If , then P(B|A) + P(A|B) equals

A. 1|4

B. 1|3

C. 5|12

C. 7|2

Answer:

Given-

Hence, the Correct option is D

Question:59

If , the P(A′|B′).P(B′|A′) is equal to

A. 5|6

B. 5|7

C. 25|42

D. 1

Answer:

Given-

Hence, the correct option is C

Question:60

If A and B are two events such that , the P(A′ ∩ B′) equals

A. 1|12

B. 3|4

C. 1|4

D. 3|16

Answer:

Given-

As we know,

P(A|B) × P(B) = P(A ∩ B) [Property of Conditional Probability]

Hence, the correct option is C

Question:61

If P(A) = 0.4, P(B) = 0.8 and P(B | A) = 0.6, then P(A ∪ B) is equal to

A. 0.24

B. 0.3

C. 0.48

D. 0.96

Answer:

Given-

P(A) = 0.4, P(B) = 0.8 and

It is known that,

Hence,

Question:62

If A and B are two events and A ≠ θ, B ≠ θ, then

A. P(A | B) = P(A).P(B)

B.

C. P(A | B).P(B | A)=1

D. P(A | B) = P(A) | P(B)

Answer:

Given-

CASE 1 : If we take option

CASE 2 : If we take option (B) i.e.

this is true, knowing that this is conditional probability.

CASE 3: If we take option

Hence, the correct option is B

Question:63

A and B are events such that P(A) = 0.4, P(B) = 0.3 and P(A ∪ B) = 0.5. Then P (B′ ∩ A) equals

A. 2|3

B. 1|2

C. 3|10

D. 1|5

Answer:

Given-

Hence, the correct option is D

Question:64

You are given that A and B are two events such that P(B)= 3|5, P(A | B) = 1|2 and P(A ∪ B) = 4|5, then P(A) equals

A. 3|10

B. 1|5

C. 1/2

D. 3|5

Answer:

Given

Hence, the correct option is C

Question:65

In Exercise 64 above, P(B | A′) is equal to

A. 1|5

B. 3|10

C. 1|2

D. 3|5

Answer:

Referring to the above solution,

Hence, the correct option is D

Question:66

If P(B) = 3|5, P(A|B) = 1|2 and P(A ∪ B) = 4|5, then P(A ∪ B)′ + P(A′ ∪ B) =

A. 1|5

B. 4|5

C. 1|2

D. 1

Answer:

Given-

Hence, the correct option is D

Question:67

Let P(A) = 7/13, P(B) = 9/13 and P(A ∩ B) = 4/13. Then P(A′|B) is equal to

A. 6/13

B. 4/13

C. 4/9

D. 5/9

Answer:

Given-

P(A) = 7/13, P(B) = 9/13 and P(A ∩ B) = 4/13

Question:68

If A and B such events that P(A) > 0 and P(B) ≠ 1, then P(A’|B’) equals

A. 1 – P(A|B)

B. 1 – P (A’|B)

C.

D. P(A’) | P(B’)

Answer:

Given-

By de Morgan's Law:

Option c is correct answer.

Question:69

If A and B are two independent events with P(A) = 3/5 and P(B) = 4/9, then P (A′ ∩ B′) equals

A.4/15

B. 8/45

C. 1/3

D. 2/9

Answer:

Question:70

If two events are independent, then

A. they must be mutually exclusive

B. the sum of their probabilities must be equal to 1

C. (A) and (B) both are correct

D. None of the above is correct

Answer:

Events which cannot happen at the same time are known as mutually exclusive events. For example: when tossing a coin, the result can either be heads or tails but cannot be both.

Events are independent if the occurrence of one event does not influence (and is not influenced by) the occurrence of the other(s).

Eg: Rolling a die and flipping a coin. The probability of getting any number on the die will not affect the probability of getting head or tail in the coin.

Therefore, if A and B events are independent, any information about A cannot tell anything about B while if they are mutually exclusive then we know if A occurs B does not occur.

Therefore, independent events cannot be mutually exclusive.

To test if probability of independent events is 1 or not:

Let A be the event of obtaining a head.

P(A) = 1/2

Let B be the event of obtaining 5 on a die.

P(B) = 1/6

Now A and B are independent events.

Hence option D is correct.

Question:71

Let A and B be two events such that P(A) = 3/8, P(B) = 5/8 and P(A ∪ B) = 3/4. Then P(A | B).P(A′ | B) is equal to

A.2/5

B. 3/8

C. 3/20

D. 6/25

Answer:

Option D is correct.

Question:72

If the events A and B are independent, then P(A ∩ B) is equal to

A. P (A) + P

B. (B) P(A) – P(B)

C. P (A) . P(B)

D. P(A) | P(B)

Answer:

We know that if events A and B are independent,

From the definition of the independent Event,

Option C is correct.

Question:73

Two events E and F are independent. If P(E) = 0.3, P(E ∪ F) = 0.5, then P(E | F)–P(F | E) equals

A. 2/7

B. 3/25

C. 1/70

D. 1/7

Answer:

Given-

P(E) = 0.3, P(E ∪ F) = 0.5

Also, E and F are independent, therefore,

P (E ∩ F)=P(E).P(F)

As we know , P(E ∪ F)=P(E)+P(F)- P(E ∩ F)

P(E ∪ F)=P(E)+P(F)- [P(E) P(F)]

Question:74

A bag contains 5 red and 3 blue balls. If 3 balls are drawn at random without replacement the probability of getting exactly one red ball is

A. 45/196

B. 135/392

C. 15/56

D. 15/29

Answer:

The Probability of getting exactly one red ball is

P(R).P(B).P(B) + P(B).P(R).P(B) + P(B).P(B).P(R)

Option C is correct.

Question:75

Refer to Question 74 above. The probability that exactly two of the three balls were red, the first ball being red, is

A. 1/3

B. 4/7

C. 15/28

D. 5/28

Answer:

Given-

A bag contains 5 red and 3 blue balls

Therefore, Total Balls in a Bag = 8

For exactly 1 red ball probability should be

3 Balls are drawn randomly then possibility for getting 1 red ball

P(E)=P(R).P(B)+P(B).P(R)

Option B is correct.

Question:76

Three persons, A, B and C, fire at a target in turn, starting with A. Their probability of hitting the target are 0.4, 0.3 and 0.2 respectively. The probability of two hits is

A. 0.024

B. 0.188

C. 0.336

D. 0.452

Answer:

Given-

Hence, Probability of two hits is 0.188

Option B is correct.

Question:77

Assume that in a family, each child is equally likely to be a boy or a girl. A family with three children is chosen at random. The probability that the eldest child is a girl given that the family has at least one girl is

A. 1/2

B. 1/3

C. 2/3

D. 4/7

Answer:

The statement can be arranged in a set as S={(B,B,B),(G,G,G),(B,G,G),(G,B,G),(G,G,B),(G,B,B),(B,G,B),(B,B,G)}

Let A be Event that a family has at least one girl, therefore,

A={(G,B,B),(B,G,B),(B,B,G),(G,G,B),(B,G,G)(G,B,G),(G,G,G)

Let B be Event that eldest child is girl then, therefore,

B={(G,B,B)(G,G,B),(G,B,G),(G,G,G)

(A ∩ B)={(G,B,B),(G,G,B),(G,B,G,)(G,G,G)

since,

Hence,

Option D is correct.

Question:78

A die is thrown and a card is selected at random from a deck of 52 playing cards. The probability of getting an even number on the die and a spade card is

A. 1/2

B. 1/4

C. 1/8

D. 3/4

Answer:

Let A be Event for getting number on dice and B be Event that a spade card is selected

Therefore,

A={2,4,6}

B={13}

Question:79

Answer:

Given-

There are total 8 balls in box.

Therefore, P(G) , Probability of green ball

P(B) , Probability of blue ball

The probability of drawing 2 green balls and one blue ball is

P(E)=P(G).P(G).P(B)+P(B).P(G).P(G)+P(G).P(B).P(G)

Question:80

Answer:

Given-

Total number of batteries: n= 8

Number of dead batteries are = 3

Therefore, Probability of dead batteries is

If two batteries are selected without replacement and tested

Then, Probability of second battery without replacement is

Required probability =

Question:81

Eight coins are tossed together. The probability of getting exactly 3 heads is

A.

B.

C.

D.

Answer:

Given-

probability distribution

Total number coin is tossed, n=8

The probability of getting head,

The probability of getting tail,

The Required probability

Question:82

Answer:

Let A be the event that the sum of numbers on the dice was less than 6

And B be the event that the sum of numbers on the dice is 3

Therefore,

A={(1,4)(4,1)(2,3)(3,2)(2,2)(1,3)(3,1)(1,2)(2,1)(1,1)

n(A)=10

B={(1,2)(2,1)

n(B)=2

Required probability =

Required probability =

Hence, the probability is

Question:83

Which one is not a requirement of a binomial distribution?

A. There are 2 outcomes for each trial

B. There is a fixed number of trials

C. The outcomes must be dependent on each other

D. The probability of success must be the same for all the trials

Answer:

In the binomial distribution, there are 2 outcomes for each trial and there is a fixed number of trials and the probability of success must be the same for all trials.

Hence option c is correct.

Question:84

Answer:

We know that

Number of cards = 52

Number of queens = 4

Therefore, Probability of queen out of 52 cards =

According to the question,

If a deck of card shuffled again with replacement, then

Probability of getting queen is ,

Therefore, The probability, that both cards are queen is ,

Hence, Probability is

Question:85

The probability of guessing correctly at least 8 out of 10 answers on a true-false type examination is

A.

B.

C.

D.

Answer:

We know that in the examination, we have only two option True and False.

Therefore, the probability of getting True is, p= 1/2

And, probability of getting False is, q=1/2

The total number of Answer in examination, n=10

The probability of guessing correctly at least 8 it means r=8,9,10

As we know, the probability distribution

Hence, Probability is

Question:86

Answer:

Given-

Total number of person n=5

Total number of swimmers among total person , r=4

Probability of not swimmer , Q=0.3

Therefore, The probability of swimmer , p=1-Q=0.7

As we know that the probability distribution

=

Hence,

Option A is correct.

Question:87

The probability distribution of a discrete random variable X is given below:

The value of k is

A. 8

B. 16

C. 32

D. 48

Answer:

Given-

Probability distribution table

As we know

Hence, the value of k is 32

Option C is correct.

Question:88

For the following probability distribution:

E(X) is equal to:

A. 0

B. –1

C. –2

D. –1.8

Answer:

Given-

Probability distribution table

Option D is correct.

Question:89

For the following probability distribution

is equal to

A. 3

B. 5

C. 7

D. 10

Answer:

Given-

Probability distribution table

Option D is correct.

Question:90

Suppose a random variable X follows the binomial distribution with parameters n and p, where 0 < p < 1. If P(x = r) / P(x = n–r) is independent of n and r, then p equals

A. 1/2

B. 1/3

C. 1/5

D. 1/7

Answer:

According to the question, this expression is independent of n and r if

Hence

Option A is correct.

Question:91

Answer:

Let A be the event that students failed in physics.

As per the question, 30% students failed in physics.

∴ P(A) = 0.30

Similarly, if we denote the event of failing in maths with B.

We get P(B) = 0.25

And probability of failing in both subjects can be represented using intersection as

P (A ∩ B) = 0.1

To find- a conditional probability of failing of student in physics given that she has failed in mathematics.

The situation can be represented mathematically as-

P(A|B) =?

Using the fundamental idea of conditional probability, we know that:

Our answers clearly match with option B

is the only correct choice.

Question:92

Answer:

Let E be the event that student ‘A’ solves the problem correctly.

∴ P(E) = 1/3

In the same way if we denote the event of ’B’ solving the problem correctly with F

We get P(F) = 1/4

Since both the events are independent.

∴ Probability that both the students solve the question correctly can be represented as-

Probability that both the students could not solve the question correctly can be represented as-

Given: probability of making a common error and both getting same answer.

If they are making an error, we can be sure that answer coming out is wrong.

Let S be the event of getting same answer.

above situation can be represented using conditional probability.

And if their answer is correct obviously, they will get same answer.

To find- the probability of getting a correct answer if they committed a common error and got the same answer.

Mathematically, i.e,

By observing our requirement and availability of equations, we can use Bayes theorem to solve this.

Using Bayes theorem, we get-

Substituting the values from above -

Our answers clearly match with option D.

∴ Option (D) is the only correct choice.

Question:93

Answer:

We can solve this using Bernoulli trials.

Here n = 5 (as we are drawing 5 pens only)

Success is defined when we get a defective pen.

Let p be the probability of success and q probability of failure.

∴ p = 10/100 = 0.1

And q = 1 – 0.1 = 0.9

To find- the probability of getting at most 1 defective pen.

Let X be a random variable denoting the probability of getting r number of defective pens.

∴ P (drawing atmost 1 defective pen) = P(X = 0) + P(X = 1)

The binomial distribution formula is:

Where:

x = total number of “successes.”

P = probability of success on an individual trial

n = number of trials

Our answer matches with option D.

∴ Option (D) is the only correct choice.

Question:94

Answer:

FALSE

Events are mutually exclusive when–

P(A∪B) = P(A) + P(B)

But as per the conditions in question, it is not necessary that they will meet the condition because it might be possible that

P(A ∩ B) ≠ 0

Events are independent when–

P(A ∩ B) = P(A)P(B)

Again P(A) > 0 and P(B)> 0 are not sufficient conditions to validate them.

Question:95

Answer:

TRUE

As A and B are independent

hence proved

Question:96

Answer:

False

If A and B are mutually exclusive, that means

P(A∪B) = P(A) + P(B)

From this equation it cannot be proved that

P(A ∩ B)= P(A)P(B).

Hence, it is a false statement.

Question:97

Answer:

False

If A and B are independent events, it means that

P(A ∩ B) = P(A)P(B)

From the equation it cannot be proved that

P(A∪B) = P(A) + P(B)

It is only possible if either P(A) or P(B) = 0, which is not given in question.

Hence, it is a false statement.

Question:98

Answer:

TRUE

If A and B are independent events it means that

P(A ∩ B) = P(A)P(B)

Thus, from the definition of independent event we say that statement is true.

Question:99

State True or False for the statements in the Exercise.

Another name for the mean of a probability distribution is expected value.

Answer:

TRUE

Mean gives the average of values and if it is related with probability or random variable it is often called expected value.

Question:100

Answer:

TRUE

If A and B are independent events, it means that

P(A ∩ B) = P(A)P(B)

P(A′ ∪ B) = P(A’) + P(B) – P(A’ ∩ B)

and P(A′ ∪ B) represents the probability of event ‘only B’ excluding common points.

Hence Proved

Question:101

Answer:

TRUE

If A and B are independent events, that means

∴ Statement is true.

Question:102

Answer:

True

The above equation means that:

we need to add to get the equal term

LHS is greater.

Question:103

Answer:

True

Let A, B,C be the occurrence of events A,B and C and A’,B’ and C’ not occurrence.

P(A) = P(B) = P(C) = p and P(A’) = P(B’) = P(C’) = 1-p

P (At least two of A, B, C occur) = P(A ∩ B ∩ C’) + P(A ∩ B’ ∩ C) + P(A’ ∩ B ∩ C) + P(A ∩ B ∩ C)

events are independent:

Hence, statement is true.

Question:106

Answer:

p = 1/10

As n = 5 {representing no. of trials}

p = probability of success

As it is a binomial distribution.

∴ probability of failure = q = 1 – p

Given-

P(X = 2) = 9.P(X = 3)

The binomial distribution formula is:

Where:

x = total number of “successes.”

P = probability of success on an individual trial

n = number of trials

using binomial distribution,

Question:107

Answer:

Variance is the mean of deviation of Random variable from its expected value.

On expanding we get the formula:

Question:108

Answer:

we know that

Given-

This implies that A and B are independent of each other.

NCERT exemplar solutions for Class 12 Maths chapter 13 provided here for the NCERT books are very useful and detailed from the point of view of aiding practice, preparation and working for Board exams as well as the JEE Main exams.

NCERT exemplar Class 12 Maths solutions chapter 13 PDF download are also available for students for extended learning. The topics covered are as follows:

- Introduction
- Conditional Probability
- Properties of conditional probability
- Multiplication theorem on probability
- Independent events
- Bayes’ Theorem
- Partition of a sample space
- Theorem of total probability
- Random Variables and its Probability Distributions
- Probability distribution of a random variable
- Mean of a random variable
- Variance of a random variable
- Bernoulli Trials and Binomial Distribution
- Bernoulli trials
- Binomial distribution

Today’s society has adopted a very practical approach to life and has understood the principle of chance in nature. Similarly, the importance of studying probability could be inferred from the basic lives of people which displays a great example at how there is a constant need to understand the possibility and extract information out of it for one’s benefit. NCERT exemplar Class 12 Math solutions chapter 13 will help students to understand each topic in a better way. Such instances of life display the necessity and importance of learning the concept of probability or possibility of any event to make important decisions in their life and how they are drawn based on chances. Such as in life you come across a lot of data which need to be interpreted, and decisions are to be extracted out of such data on the basis of probable thinking which necessitates the learning of probabilistic thinking.

NCERT exemplar Class 12 Math solutions chapter 13 are easy to grasp and can be helpful in scoring well in the exams. It is not only theoretical/academic oriented but provides a great exposure regarding the nature of chance and variation in life when you encounter such situations in real life. It also develops and exhibits real-life situations teaching the art of risk analysis and management, providing a better approach towards understanding risks in real life and displaying relative chances of events in one’s life. This topic also helps in scientific reasoning and also has its imprint in various fields such as engineering, mathematician, research analyst, etc. Probability also supports statistics and helps improve and analyze data-driven decision-making skills for real life.

- Class 12 Math NCERT exemplar solutions chapter 13 displays important concept including that of the conditional probability of any event given that another event has already occurred which will be helpful in understanding different theorems relating to probability, multiplication rule of probability, independence of events and discrete probability distribution also known as Binomial distribution.
- NCERT exemplar Class 12 Math solutions chapter 13 also highlights the concept of the random variable, its probability distribution and also the mean and variance of any probability distribution.

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Chapter 12 | |

Chapter 13 | Probability |

- NCERT Solution for Class 12 Physics
- NCERT Solution for Class 12 Chemistry
- NCERT Solution for Class 12 Maths
- NCERT Solution for Class 12 Biology

- NCERT Notes for Class 12 Physics
- NCERT Notes for Class 12 Chemistry
- NCERT Notes for Class 12 Maths
- NCERT Notes for Class 12 Biology

1. How can probability be useful?

Understanding probability can be highly effective in solving real life problems and also in understanding higher education topics.

2. How many questions are included in solutions?

All these questions that are mentioned in the main exercise after the chapter and in the additional question section are solved in NCERT exemplar Class 12 Maths solutions chapter 13.

3. Are these questions helpful in board exams?

Yes, these solutions are highly useful for board exams as it helps in learning the steps and the way the CBSE wants students to solve questions.

4. Are these questions useful in competitive exams?

Yes, these NCERT exemplar Class 12 Maths solutions chapter 13 pdf download are useful in entrance exams for engineering and also are helpful for many other competitive exams.

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5 minHave a question related to CBSE Class 12th ?

Hello aspirant,

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Thank you

Hope this information helps you.

hello,

Yes you can appear for the compartment paper again since CBSE gives three chances to a candidate to clear his/her exams so you still have two more attempts. However, you can appear for your improvement paper for all subjects but you cannot appear for the ones in which you have failed.

I hope this was helpful!

Good Luck

Hello dear,

If you was not able to clear 1st compartment and now you giving second compartment so YES, you can go for your improvement exam next year but if a student receives an improvement, they are given the opportunity to retake the boards as a private candidate the following year, but there are some requirements. First, the student must pass all of their subjects; if they received a compartment in any subject, they must then pass the compartment exam before being eligible for the improvement.

As you can registered yourself as private candidate for giving your improvement exam of 12 standard CBSE(Central Board of Secondary Education).For that you have to wait for a whole year which is bit difficult for you.

Positive side of waiting for whole year is you have a whole year to preparing yourself for your examination. You have no distraction or something which may causes your failure in the exams. In whole year you have to stay focused on your 12 standard examination for doing well in it. By this you get a highest marks as a comparison of others.

Believe in Yourself! You can make anything happen

All the very best.

Hello Student,

I appreciate your Interest in education. See the improvement is not restricted to one subject or multiple subjects and we cannot say if improvement in one subject in one year leads to improvement in more subjects in coming year.

You just need to have a revision of all subjects what you have completed in the school. have a revision and practice of subjects and concepts helps you better.

All the best.

Hi,

You just need to give the exams for the concerned two subjects in which you have got RT. There is no need to give exam for all of your subjects, you can just fill the form for the two subjects only.

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Remote sensing scientists use remote sensing technology to support scientists in fields such as community planning, flight planning or the management of natural resources. Analysing data collected from aircraft, satellites or ground-based platforms using statistical analysis software, image analysis software or Geographic Information Systems (GIS) is a significant part of their work. Do you want to learn how to become remote sensing technician? There's no need to be concerned; we've devised a simple remote sensing technician career path for you. Scroll through the pages and read.

3 Jobs Available

Budget analysis, in a nutshell, entails thoroughly analyzing the details of a financial budget. The budget analysis aims to better understand and manage revenue. Budget analysts assist in the achievement of financial targets, the preservation of profitability, and the pursuit of long-term growth for a business. Budget analysts generally have a bachelor's degree in accounting, finance, economics, or a closely related field. Knowledge of Financial Management is of prime importance in this career.

4 Jobs Available

The invention of the database has given fresh breath to the people involved in the data analytics career path. Analysis refers to splitting up a whole into its individual components for individual analysis. Data analysis is a method through which raw data are processed and transformed into information that would be beneficial for user strategic thinking.

Data are collected and examined to respond to questions, evaluate hypotheses or contradict theories. It is a tool for analyzing, transforming, modeling, and arranging data with useful knowledge, to assist in decision-making and methods, encompassing various strategies, and is used in different fields of business, research, and social science.

3 Jobs Available

An underwriter is a person who assesses and evaluates the risk of insurance in his or her field like mortgage, loan, health policy, investment, and so on and so forth. The underwriter career path does involve risks as analysing the risks means finding out if there is a way for the insurance underwriter jobs to recover the money from its clients. If the risk turns out to be too much for the company then in the future it is an underwriter who will be held accountable for it. Therefore, one must carry out his or her job with a lot of attention and diligence.

3 Jobs Available

A Product Manager is a professional responsible for product planning and marketing. He or she manages the product throughout the Product Life Cycle, gathering and prioritising the product. A product manager job description includes defining the product vision and working closely with team members of other departments to deliver winning products.

3 Jobs Available

Individuals in the operations manager jobs are responsible for ensuring the efficiency of each department to acquire its optimal goal. They plan the use of resources and distribution of materials. The operations manager's job description includes managing budgets, negotiating contracts, and performing administrative tasks.

3 Jobs Available

Individuals who opt for a career as a stock analyst examine the company's investments makes decisions and keep track of financial securities. The nature of such investments will differ from one business to the next. Individuals in the stock analyst career use data mining to forecast a company's profits and revenues, advise clients on whether to buy or sell, participate in seminars, and discussing financial matters with executives and evaluate annual reports.

2 Jobs Available

A Researcher is a professional who is responsible for collecting data and information by reviewing the literature and conducting experiments and surveys. He or she uses various methodological processes to provide accurate data and information that is utilised by academicians and other industry professionals. Here, we will discuss what is a researcher, the researcher's salary, types of researchers.

2 Jobs Available

Welding Engineer Job Description: A Welding Engineer work involves managing welding projects and supervising welding teams. He or she is responsible for reviewing welding procedures, processes and documentation. A career as Welding Engineer involves conducting failure analyses and causes on welding issues.

5 Jobs Available

A career as Transportation Planner requires technical application of science and technology in engineering, particularly the concepts, equipment and technologies involved in the production of products and services. In fields like land use, infrastructure review, ecological standards and street design, he or she considers issues of health, environment and performance. A Transportation Planner assigns resources for implementing and designing programmes. He or she is responsible for assessing needs, preparing plans and forecasts and compliance with regulations.

3 Jobs Available

Individuals who opt for a career as an environmental engineer are construction professionals who utilise the skills and knowledge of biology, soil science, chemistry and the concept of engineering to design and develop projects that serve as solutions to various environmental problems.

2 Jobs Available

A Safety Manager is a professional responsible for employee’s safety at work. He or she plans, implements and oversees the company’s employee safety. A Safety Manager ensures compliance and adherence to Occupational Health and Safety (OHS) guidelines.

2 Jobs Available

A Conservation Architect is a professional responsible for conserving and restoring buildings or monuments having a historic value. He or she applies techniques to document and stabilise the object’s state without any further damage. A Conservation Architect restores the monuments and heritage buildings to bring them back to their original state.

2 Jobs Available

A Structural Engineer designs buildings, bridges, and other related structures. He or she analyzes the structures and makes sure the structures are strong enough to be used by the people. A career as a Structural Engineer requires working in the construction process. It comes under the civil engineering discipline. A Structure Engineer creates structural models with the help of computer-aided design software.

2 Jobs Available

**Highway Engineer Job Description: **A Highway Engineer is a civil engineer who specialises in planning and building thousands of miles of roads that support connectivity and allow transportation across the country. He or she ensures that traffic management schemes are effectively planned concerning economic sustainability and successful implementation.

2 Jobs Available

Are you searching for a Field Surveyor Job Description? A Field Surveyor is a professional responsible for conducting field surveys for various places or geographical conditions. He or she collects the required data and information as per the instructions given by senior officials.

2 Jobs Available

Orthotists and Prosthetists are professionals who provide aid to patients with disabilities. They fix them to artificial limbs (prosthetics) and help them to regain stability. There are times when people lose their limbs in an accident. In some other occasions, they are born without a limb or orthopaedic impairment. Orthotists and prosthetists play a crucial role in their lives with fixing them to assistive devices and provide mobility.

6 Jobs Available

A career in pathology in India is filled with several responsibilities as it is a medical branch and affects human lives. The demand for pathologists has been increasing over the past few years as people are getting more aware of different diseases. Not only that, but an increase in population and lifestyle changes have also contributed to the increase in a pathologist’s demand. The pathology careers provide an extremely huge number of opportunities and if you want to be a part of the medical field you can consider being a pathologist. If you want to know more about a career in pathology in India then continue reading this article.

5 Jobs Available

Gynaecology can be defined as the study of the female body. The job outlook for gynaecology is excellent since there is evergreen demand for one because of their responsibility of dealing with not only women’s health but also fertility and pregnancy issues. Although most women prefer to have a women obstetrician gynaecologist as their doctor, men also explore a career as a gynaecologist and there are ample amounts of male doctors in the field who are gynaecologists and aid women during delivery and childbirth.

4 Jobs Available

The audiologist career involves audiology professionals who are responsible to treat hearing loss and proactively preventing the relevant damage. Individuals who opt for a career as an audiologist use various testing strategies with the aim to determine if someone has a normal sensitivity to sounds or not. After the identification of hearing loss, a hearing doctor is required to determine which sections of the hearing are affected, to what extent they are affected, and where the wound causing the hearing loss is found. As soon as the hearing loss is identified, the patients are provided with recommendations for interventions and rehabilitation such as hearing aids, cochlear implants, and appropriate medical referrals. While audiology is a branch of science that studies and researches hearing, balance, and related disorders.

3 Jobs Available

An oncologist is a specialised doctor responsible for providing medical care to patients diagnosed with cancer. He or she uses several therapies to control the cancer and its effect on the human body such as chemotherapy, immunotherapy, radiation therapy and biopsy. An oncologist designs a treatment plan based on a pathology report after diagnosing the type of cancer and where it is spreading inside the body.

3 Jobs Available

Are you searching for an ‘Anatomist job description’? An Anatomist is a research professional who applies the laws of biological science to determine the ability of bodies of various living organisms including animals and humans to regenerate the damaged or destroyed organs. If you want to know what does an anatomist do, then read the entire article, where we will answer all your questions.

2 Jobs Available

For an individual who opts for a career as an actor, the primary responsibility is to completely speak to the character he or she is playing and to persuade the crowd that the character is genuine by connecting with them and bringing them into the story. This applies to significant roles and littler parts, as all roles join to make an effective creation. Here in this article, we will discuss how to become an actor in India, actor exams, actor salary in India, and actor jobs.

4 Jobs Available

Individuals who opt for a career as acrobats create and direct original routines for themselves, in addition to developing interpretations of existing routines. The work of circus acrobats can be seen in a variety of performance settings, including circus, reality shows, sports events like the Olympics, movies and commercials. Individuals who opt for a career as acrobats must be prepared to face rejections and intermittent periods of work. The creativity of acrobats may extend to other aspects of the performance. For example, acrobats in the circus may work with gym trainers, celebrities or collaborate with other professionals to enhance such performance elements as costume and or maybe at the teaching end of the career.

3 Jobs Available

Career as a video game designer is filled with excitement as well as responsibilities. A video game designer is someone who is involved in the process of creating a game from day one. He or she is responsible for fulfilling duties like designing the character of the game, the several levels involved, plot, art and similar other elements. Individuals who opt for a career as a video game designer may also write the codes for the game using different programming languages.

Depending on the video game designer job description and experience they may also have to lead a team and do the early testing of the game in order to suggest changes and find loopholes.

3 Jobs Available

Radio Jockey is an exciting, promising career and a great challenge for music lovers. If you are really interested in a career as radio jockey, then it is very important for an RJ to have an automatic, fun, and friendly personality. If you want to get a job done in this field, a strong command of the language and a good voice are always good things. Apart from this, in order to be a good radio jockey, you will also listen to good radio jockeys so that you can understand their style and later make your own by practicing.

A career as radio jockey has a lot to offer to deserving candidates. If you want to know more about a career as radio jockey, and how to become a radio jockey then continue reading the article.

3 Jobs Available

The word “choreography" actually comes from Greek words that mean “dance writing." Individuals who opt for a career as a choreographer create and direct original dances, in addition to developing interpretations of existing dances. A Choreographer dances and utilises his or her creativity in other aspects of dance performance. For example, he or she may work with the music director to select music or collaborate with other famous choreographers to enhance such performance elements as lighting, costume and set design.

2 Jobs Available

A career as social media manager involves implementing the company’s or brand’s marketing plan across all social media channels. Social media managers help in building or improving a brand’s or a company’s website traffic, build brand awareness, create and implement marketing and brand strategy. Social media managers are key to important social communication as well.

2 Jobs Available

Photography is considered both a science and an art, an artistic means of expression in which the camera replaces the pen. In a career as a photographer, an individual is hired to capture the moments of public and private events, such as press conferences or weddings, or may also work inside a studio, where people go to get their picture clicked. Photography is divided into many streams each generating numerous career opportunities in photography. With the boom in advertising, media, and the fashion industry, photography has emerged as a lucrative and thrilling career option for many Indian youths.

2 Jobs Available

An individual who is pursuing a career as a producer is responsible for managing the business aspects of production. They are involved in each aspect of production from its inception to deception. Famous movie producers review the script, recommend changes and visualise the story.

They are responsible for overseeing the finance involved in the project and distributing the film for broadcasting on various platforms. A career as a producer is quite fulfilling as well as exhaustive in terms of playing different roles in order for a production to be successful. Famous movie producers are responsible for hiring creative and technical personnel on contract basis.

2 Jobs Available

In a career as a copywriter, one has to consult with the client and understand the brief well. A career as a copywriter has a lot to offer to deserving candidates. Several new mediums of advertising are opening therefore making it a lucrative career choice. Students can pursue various copywriter courses such as Journalism, Advertising, Marketing Management. Here, we have discussed how to become a freelance copywriter, copywriter career path, how to become a copywriter in India, and copywriting career outlook.

5 Jobs Available

In a career as a vlogger, one generally works for himself or herself. However, once an individual has gained viewership there are several brands and companies that approach them for paid collaboration. It is one of those fields where an individual can earn well while following his or her passion.

Ever since internet costs got reduced the viewership for these types of content has increased on a large scale. Therefore, a career as a vlogger has a lot to offer. If you want to know more about the Vlogger eligibility, roles and responsibilities then continue reading the article.

3 Jobs Available

For publishing books, newspapers, magazines and digital material, editorial and commercial strategies are set by publishers. Individuals in publishing career paths make choices about the markets their businesses will reach and the type of content that their audience will be served. Individuals in book publisher careers collaborate with editorial staff, designers, authors, and freelance contributors who develop and manage the creation of content.

3 Jobs Available

Careers in journalism are filled with excitement as well as responsibilities. One cannot afford to miss out on the details. As it is the small details that provide insights into a story. Depending on those insights a journalist goes about writing a news article. A journalism career can be stressful at times but if you are someone who is passionate about it then it is the right choice for you. If you want to know more about the media field and journalist career then continue reading this article.

3 Jobs Available

Individuals in the editor career path is an unsung hero of the news industry who polishes the language of the news stories provided by stringers, reporters, copywriters and content writers and also news agencies. Individuals who opt for a career as an editor make it more persuasive, concise and clear for readers. In this article, we will discuss the details of the editor's career path such as how to become an editor in India, editor salary in India and editor skills and qualities.

3 Jobs Available

Individuals who opt for a career as a reporter may often be at work on national holidays and festivities. He or she pitches various story ideas and covers news stories in risky situations. Students can pursue a BMC (Bachelor of Mass Communication), B.M.M. (Bachelor of Mass Media), or MAJMC (MA in Journalism and Mass Communication) to become a reporter. While we sit at home reporters travel to locations to collect information that carries a news value.

2 Jobs Available

Are you searching for a Corporate Executive job description? A Corporate Executive role comes with administrative duties. He or she provides support to the leadership of the organisation. A Corporate Executive fulfils the business purpose and ensures its financial stability. In this article, we are going to discuss how to become corporate executive.

2 Jobs Available

A multimedia specialist is a media professional who creates, audio, videos, graphic image files, computer animations for multimedia applications. He or she is responsible for planning, producing, and maintaining websites and applications.

2 Jobs Available

5 Jobs Available

A quality controller plays a crucial role in an organisation. He or she is responsible for performing quality checks on manufactured products. He or she identifies the defects in a product and rejects the product.

A quality controller records detailed information about products with defects and sends it to the supervisor or plant manager to take necessary actions to improve the production process.

3 Jobs Available

3 Jobs Available

A QA Lead is in charge of the QA Team. The role of QA Lead comes with the responsibility of assessing services and products in order to determine that he or she meets the quality standards. He or she develops, implements and manages test plans.

2 Jobs Available

2 Jobs Available

The Process Development Engineers design, implement, manufacture, mine, and other production systems using technical knowledge and expertise in the industry. They use computer modeling software to test technologies and machinery. An individual who is opting career as Process Development Engineer is responsible for developing cost-effective and efficient processes. They also monitor the production process and ensure it functions smoothly and efficiently.

2 Jobs Available

An AWS Solution Architect is someone who specializes in developing and implementing cloud computing systems. He or she has a good understanding of the various aspects of cloud computing and can confidently deploy and manage their systems. He or she troubleshoots the issues and evaluates the risk from the third party.

4 Jobs Available

An Azure Administrator is a professional responsible for implementing, monitoring, and maintaining Azure Solutions. He or she manages cloud infrastructure service instances and various cloud servers as well as sets up public and private cloud systems.

4 Jobs Available

Careers in computer programming primarily refer to the systematic act of writing code and moreover include wider computer science areas. The word 'programmer' or 'coder' has entered into practice with the growing number of newly self-taught tech enthusiasts. Computer programming careers involve the use of designs created by software developers and engineers and transforming them into commands that can be implemented by computers. These commands result in regular usage of social media sites, word-processing applications and browsers.

3 Jobs Available

3 Jobs Available

Individuals in the information security manager career path involves in overseeing and controlling all aspects of computer security. The IT security manager job description includes planning and carrying out security measures to protect the business data and information from corruption, theft, unauthorised access, and deliberate attack

3 Jobs Available

An Automation Test Engineer job involves executing automated test scripts. He or she identifies the project’s problems and troubleshoots them. The role involves documenting the defect using management tools. He or she works with the application team in order to resolve any issues arising during the testing process.

2 Jobs Available

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