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RD Sharma Class 12 Exercise 30.2 Probability Solutions Maths-Download PDF Online

RD Sharma Class 12 Exercise 30.2 Probability Solutions Maths-Download PDF Online

Edited By Lovekush kumar saini | Updated on Jan 25, 2022 04:01 PM IST

The RD Sharma class 12 solution Probability exercise 30.2 is a tricky chapter to be solved by any student. The RD Sharma class 12th exercise 30.2 starts with the explanation of the basic concepts of the chapter and then moves on to introducing the challenging questions to prepare the students in every way out to score high in the maths exam. The Class 12 RD Sharma chapter 30 exercise 30.2 solution consists of a total of 16 questions to give a brief of the essential concepts of the chapter that are mentioned below-

This Story also Contains
  1. RD Sharma Class 12 Solutions Chapter30 Probability - Other Exercise
  2. Probability Excercise: 30.2
  3. RD Sharma Chapter-wise Solutions
  • The axiomatic approach of Probability

  • The classical theory of Probability

  • Conditional Probability

  • Permutation and Combination

  • With replacement and without replacement type questions

  • Union and Intersection

  • Question based on Cards and balls

RD Sharma Class 12 Solutions Chapter30 Probability - Other Exercise

Probability Excercise: 30.2

Probability exercise 30.2 question 1

Answer:
1221
Hint:
Use formula of probability, where
P(AB)=P(A)×P(BA)
Given, a pack of 52 cards.
Solution: So Let, A = A king in first trial
B = A king in second trial.
Now, Total cards are 52.
P(A)= No. of elements of A  Total no. of elements =452=113
Now
P(BA)= A consecutive king without replacement after first trial.= No. of kings in cards  Total number of cards after excluding one king =351 (After trail cards will be and there will be kings) =117
Now,
 Required Probability =P(AB)=P(A)×P(BA)=113×117=1221

Probability exercise 30.2 question 2

Answer:
1270725
Hint:
 Probability = No. of favourable outcomes  Total no. of outcomes 
Solution:
Let, A = An ace in the first draw.
B = An ace in the 2nd draw.
C = An ace in the 3rd draw.
D = An ace in the 4th draw.
 Probability of A =425=113P(BA)=351=117P(CAB)=250=125P(DABC)=149
 Required Probability =P(ABCD)=P(A)×P(BA)×P(CAB)×P(DABC)=113×117×125×149=1270725

Probability exercise 30.2 question 3

Answer:
722
Hint:
 Probability = No. of favourable outcomes  Total no. of outcomes 
Let, A = White ball in first draw.
B = White ball in second draw.
Solution:
Now,
P(A)=712P(BA)=611 Required Probability =P(AB)=P(A)×P(BA)=712×611=42132=722

Probability exercise 30.2 question 4

Answer:
1150
Hint:
Consider events for tickets and use probability formula
Given, 12 even number between 1 to 25
Solution:
Let,
A = An even number ticket in the first draw
B = An even number ticket in the second draw
Now,
P(A)=1225P(BA)=1125 Required Probability =P(AB)=P(A)×P(BA)=1525×1124=1150

Probability exercise 30.2 question 5

Answer:
11850
Hint:
To find probability that each time it is card of a spade, calculate the events and take ratio with total events
Let, A = An ace in the 1st draw
B = An ace in the 2nd draw
C = Getting an ace in the 3rd draw.
Now,
P(A)=1352=14P(BA)=1251=417P(CBA)=1150 Required Probability =P(ABC)=P(A)×P(BA)×CBA=14×417×1150=11850


Probability exercise 30.2 question 6(i)

Answer:
1221
Hint:
 Probability = No. of event favourable outcomes  Total no. of outcomes 
Given events
Let A=A king in 1st draw
B= A king in 2nd draw
Solution:
P(A)=452=113P(BA)=351=117 Required Probability =P(AB)=P(A)×P(BA)=113×117=1221

Probability exercise 30.2 question 6(ii)

Answer:
4663
Hint:
 Probability = No. of event favourable outcomes  Total no. of outcomes 
Given Events
Let A= A king in 1st draw
B= An ace in the 2nd draw
Solution:
P(A)=452=113P(BA)=451 Required Probability =P(AB)=P(A)×P(BA)=113×451=4663

Probability exercise 30.2 question 6(iii)

Answer:
25204
Hint:
 Probability = No. of event favourable outcomes  Total no. of outcomes 
Given, events as follows
Let C=A heart in the 1st throw
D=A red card in the 2nd throw
Now,
P(C)=1352=14P(CD)=2551 Required Probability =P(CD)=P(C)×P(CD)=14×2551=25204

Probability exercise 30.2 question 7

Answer:
519
Hint:
Just take events and apply probability.
Given,
A=An even number in the 1st draw
B= An odd number in the 2nd draw
Now,
P(A)=1020=12P(BA)=1019 Required Probability =P(AB)=P(A)×P(BA)=12×1019=519

Probability exercise 30.2 question 8

Answer:
1522
Hint:
 Probability = No. of event favourable outcomes  Total no. of outcomes 
Given events
A=A white or red ball in the 1st draw.
B=A white or red ball in the 2nd draw.
Now,
P(A)=712P(BA)=611P(AB)=P(A)×P(BA)=712×611=722 Required Probability =1P(AB)=1722=1522

Probability exercise 30.2 question 9

Answer:
865
Hint:
 Probability = No. of event favourable outcomes  Total no. of outcomes 
Given, events
Let A=A white or black ball in the 1st draw
B=A white or black ball in the 2nd draw
C=A white or black ball in the 3rd draw
Now,
P(A)=815P(BA)=714=12P(CBA)=613 Required Probability =P(ABC)=P(A)×P(BA)×P(CBA)=815×12×613=865

Probability exercise 30.2 question 10

Answer:
13204
Hint:
 Probability = No. of event favourable outcomes  Total no. of outcomes 
Given, A=A heart in the 1st draw
B=A diamond in the 2nd draw
P(A)=1352=14P(BA)=1351 Required Probability =P(AB)=P(A)×P(BA)=14×1351=13204

Probability exercise 30.2 question 11

Answer:
37
Hint:
 Probability = No. of event favourable outcomes  Total no. of outcomes 
Given, events
A=A black ball in the 1st draw
B=A black ball in the 2nd draw
Now,
P(A)=1015=23P(BA)=914 Required Probability =P(AB)=P(A)×P(BA)=23×914=37

Probability exercise 30.2 question 12

Answer:
25525
Hint:
 Probability =P(ABC)=P(A)×P(BA)×P(CBA)
Given, events
A=A king in the 1st draw
B=A king in the 2nd draw
P(A)=452=113P(BA)=351=117P(CBA)=450=225 Required Probability =P(ABC)=P(A)×P(BA)×P(CBA)=113×117×225=25525

Probability exercise 30.2 question 13

Answer:
4491
Hint:
 Probability = No. of event favourable outcomes  Total no. of outcomes 
Given,
A=A good orange in the 1st draw
B=A good orange in the 2nd draw
C=A good orange in the 3rd draw
Now,
P(A)=1215=45P(BA)=1114P(CBA)=1013 Required Probability =P(ABC)=P(A)×P(BA)×P(CBA)=45×1114×1013=440910=4491

Probability exercise 30.2 question 14

Answer:
124
Hint:
 Probability = No. of event favourable outcomes  Total no. of outcomes 
Given,
A=A white ball in the 1st draw
B=A black ball in the 2nd draw
C=A red ball in the 3rd draw
P(A)=416=14P(BA)=715P(CBA)=514 Required Probability =P(ABC)=P(A)×P(BA)×P(CBA)=14×715×514=124


The RD Sharma class 12th exercise 30.2 is trusted by thousands of students in the entire country and recommended by teachers when it comes to preparing for the board examination. The students who have passed the board examinations will tell you how beneficial the RD Sharma class 12 solution chapter 30 exercise 30.2 has been in the crucial moment for the preparation. RD Sharma solutions The experts in the field of mathematics design the special questions in the RD Sharma class 12th exercise 30.2 in an alternate way providing tips to the students that help them solve the toughest questions with ease and consume less time.

The RD Sharma class 12 chapter 30 exercise 30.2 should be practiced thoroughly to ensure that you will score high marks in the board exams, as most of the questions asked in exams are based on the questions provided in the RD Sharma class 12th exercise 30.2. Teachers refer to this solution for assigning homework, so it helps solve homework as well.

The RD Sharma class 12th exercise 30.2 can be downloaded from the website of the Career360 through any device by any student residing in any part of the country. Also, the RD Sharma class 12th exercise 30.2 can be accessed free of cost as the Career360 website doesn't ask for any penny to provide the online PDFs of the RD Sharma solution, so don't waste any more time and grab your solutions asap.

RD Sharma Chapter-wise Solutions

Frequently Asked Questions (FAQs)

1. What is the cost of the RD Sharma class 12 chapter 30 solution?

It is available free of cost on the official website of Career360, and you can download it online without visiting any offline store.

2. Is Probability an important chapter for the exam?

Yes, every chapter is important in its own way for the preparation of the board exam. So practice it thoroughly to score high.

3. Can I take the help of the RD Sharma class 12th exercise 30.2 for solving homework?

Yes, it is beneficial to take the help of the RD Sharma solution to solve homework as it gives alternate ways to solve questions to save time.

4. Are the RD Sharma solutions of the updated version?

Yes, the solutions are updated yearly to correspond with the syllabus of NCERT to prepare students for any public exam.

5. Are all the materials of the RD Sharma available online?

Yes, it is available on the Career360 website; download the PDFs of all the RD Sharma chapters online.

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