NCERT solution Class 12 probability

NCERT Solutions

Class 12 Maths

Probability

NCERT Questions

Ex.13.1 Q.1
Ex.13.1 Q.2
Ex.13.1 Q.3
Ex.13.1 Q.4
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Ex.13.1 Q.16
Ex.13.1 Q.17
Ex.13.2 Q.1
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Ex.13.2 Q.18
Ex.13.3 Q.1
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Ex.13.3 Q.14
Ex.13.4 Q.1
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Ex.13.4 Q.3
Ex.13.4 Q.3
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Ex.13.4 Q.8
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Ex.13.4 Q.13
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Ex.13.4 Q.15
Ex.13.4 Q.16
Ex.13.4 Q.17
Ex.13.5 Q.1
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Ex.13.5 Q.3
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Ex.13.5 Q.5
Ex.13.5 Q.6
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Ex.13.5 Q.11
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Ex.13.5 Q.13
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Ex.13.5 Q.15
Ex.Misc.Q.1
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Ex.Misc.Q.3
Ex.MIsc. Q.5
Ex.Misc.Q.6
Ex.Misc.Q.7
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Ex.10.1 Q.1
Ex.10.1 Q.3
Ex.13.1 Q.1
Ex.13.1 Q.2
Ex.13.1 Q.3
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Ex.13.1 Q.13
Ex.13.1 Q.14
Ex.13.1 Q.14
Ex.13.1 Q.16
Ex.13.1 Q.17
Ex.13.2 Q.1
Ex.13.2 Q.2
Ex.13.2 Q.3
Ex.13.2 Q.4
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Ex.13.2 Q.17
Ex.13.2 Q.18
Ex.13.3 Q.1
Ex.13.3 Q.2
Ex.13.3 Q.3
Ex.13.3 Q.4
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Ex.13.3 Q.6
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Ex.13.3 Q.13
Ex.13.3 Q.14
Ex.13.4 Q.1
Ex.13.4 Q.2
Ex.13.4 Q.3
Ex.13.4 Q.4
Ex.13.4 Q.5
Ex.13.4 Q.6
Ex.13.4 Q.7
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Ex.13.4 Q.13
Ex.13.4 Q.14
Ex.13.4 Q.15
Ex.13.4 Q.16
Ex.13.4 Q.17
Ex.13.5 Q.1
Ex.13.5 Q.2
Ex.13.5 Q.1
Ex.13.5 Q.4
Ex.13.5 Q.5
Ex.13.5 Q.6
Ex.13.5 Q.7
Ex.13.5 Q.8
Ex.13.5 Q.9
Ex.13.5 Q.10
Ex.13.5 Q.11
Ex.13.5 Q.12
Ex.13.5 Q.13
Ex.13.5 Q.14
Ex.13.5 Q.15
Ex.Misc. Q.1
Ex.Misc. Q.2
Ex.Misc. Q.3
Ex.Misc.Q.4
Ex.Misc. Q.5
Ex.Misc.Q.6
Ex.Misc.Q.7
Ex.Misc. Q.8
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Ex.Misc.Q.10
Ex.Misc.Q.11
Ex.Misc.Q.12
Ex.Misc.Q.13
Ex.Misc.Q.14
Ex.Misc.Q.15
Ex.Misc.Q.16
Ex.Misc.Q.17
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Ex.Misc.Q.19

Ex.13.4 Q.17

Suppose that two cards are drawn at random from a deck of cards. Let X be the number of aces obtained. Then the value of E(X) is

(A)

Let X denote the number of aces obtained. Therefore, X can take any of the values of 0, 1, or 2.

In a deck of 52 cards, 4 cards are aces. Therefore, there are 48 non-ace cards.

So, P(X = 0) = P(0 ace and 2 non-ace cards) = (⁴C₀ × ⁴⁸C₂)/⁵²C₂

P(X = 1) = P(1 ace and 1 non-ace cards) = (⁴C₂ × ⁴⁸C₀)/⁵²C₂

P(X = 2) = P(2 ace and 0 non- ace cards) = (⁴C₂ × ⁴⁸C₀)/⁵²C₂

Thus, the probability distribution is as follows:

X	0	1	2
P(X)

Then, E(X) = Σ X_i × P(X_i)

Let X denote the number of aces obtained. Therefore, X can take any of the values of 0, 1, or 2.

In a deck of 52 cards, 4 cards are aces. Therefore, there are 48 non-ace cards.

So, P(X = 0) = P(0 ace and 2 non-ace cards) = (⁴C₀ × ⁴⁸C₂)/⁵²C₂

P(X = 1) = P(1 ace and 1 non-ace cards) = (⁴C₂ × ⁴⁸C₀)/⁵²C₂

P(X = 2) = P(2 ace and 0 non- ace cards) = (⁴C₂ × ⁴⁸C₀)/⁵²C₂

Thus, the probability distribution is as follows:

X	0	1	2
P(X)

Then, E(X) = Σ X_i × P(X_i)

Therefore, the correct answer is option D.

Probability Class 12 Maths NCERT Chapter 13 NCERT Solutions

Solving NCERT textbook questions of Probability Class 12 Maths NCERT Chapter 13 is a crucial step to gauge your understanding of the concepts. It provides you with an opportunity to assess your grasp of the material after learning each chapter. We have already provided solutions to all the exercises in the NCERT textbook of Class 12 Maths Probability, available in various formats such as video, text, and downloadable PDFs. If you need assistance with any of the NCERT questions or any other academic topics, feel free to ask in the “Ask Questions” section of LearnoHub.com.

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How to make the most of NCERT Solutions module of Probability Class 12 Maths NCERT Chapter 13 on LearnoHub.com?

Using the NCERT solutions module effectively can significantly enhance your learning experience and understanding of the subject matter. Here's a step-by-step guide on how to make the most out of the NCERT solutions:

Attempt to Solve Yourself: Before looking at the solution, try to solve or answer the question independently. This practice will help you test your knowledge and problem-solving skills.
Match with Our Solution: After attempting the question, compare your result and approach with the provided NCERT solution. This step allows you to identify any mistakes or gaps in your understanding.
Refer to Solution Video (if needed): If you encounter difficulty or are unable to solve a question, you can refer to the NCERT solution video on Probability Class 12 Maths NCERT Chapter 13 to get a step-by-step explanation of the correct approach. Pay attention to the reasoning and methodology used in the video.
Solve Again Yourself: Once you've understood the solution, attempt to solve the question again by yourself. Repetition and practice solidify your understanding of the concepts and improve retention.
Avoid Memorization: Focus on understanding the underlying principles and concepts behind each solution. Avoid simply memorizing the solutions, as this may hinder your ability to apply the knowledge in different contexts.
Attempt All Questions: Make an effort to attempt all the questions provided in the NCERT textbook. This comprehensive practice will help you cover the entire syllabus and reinforce your understanding across various topics. You may refer Class 12 Maths Probability Online Test on LearnoHub.com.
Review and Revise: Regularly review the concepts and solutions you've learned. Revise your notes and practice regularly to retain the knowledge effectively. You may refer Probability Class 12 Maths Notes.
Ask for Clarification: If you encounter any doubts or have questions about the solutions, don't hesitate to seek clarification on the “Ask Questions” section of LearnoHub.com. Probability Class 12 Maths NCERT Chapter 13 Questions.

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NCERT solutions of Probability Class 12 Maths - all questions of all exercises are solved. Q. No. 1, 2, 3, and so on. Complete explanation and detailed solution makes them the best videos for Class 12 Maths Probability NCERT solutions.