AP EAMCET Probability Weightage 2025
AP EAMCET Probability Weightage 2025 is expected to be 3.75% contributing 3- 4 valuable marks in the Mathematics section. Mastering core topics like Conditional Probability, Bayes' Theorem, and Binomial Distribution can significantly elevate your score in AP EAMCET 2025.
AP EAMCET Probability Weightage 2025
is expected to be 3.75% in the Mathematics section. This chapter holds 3 to 4 marks in the AP EAMCET Mathematics section. These marks can make a significant difference in the overall ranking, as AP EAMCET is highly competitive. The Probability chapter in the AP EAMCET exam covers fundamental concepts such as basic probability concepts, Conditional Probability, Bayes' Theorem, Permutations and Combinations, Independent and Dependent Events, Random Variables and Distributions, and Binomial Probability Distribution. These topics equip students with essential problem-solving skills for analyzing events, calculating probabilities, and understanding probability distributions in various scenarios. In this article, you will be able to understand the AP EAMCET Probability Weightage 2025 along with its concepts that will help you to prepare well for the exam.
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has been live. Students can fill and submit the application form without a late fee untill April 24, 2025. According to the official schedule, AP EAPCET 2025 Agriculture & Pharmacy exam will be held on
May 19 and 20, 2025
and the AP EAMCET 2025 Engineering exam will be conducted from
May 21 to 27, 2025
.
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Expected AP EAMCET Probability Weightage 2025
Candidates who are planning to appear for AP EAMCET 2025 must note that the expected AP EAMCET 2025 Probability Weightage is 3.75% of the Mathematics total score. Test takers can expect around 3 to 4 questions from this chapter. Students are advised to master all the topics in the AP EAMCET Probability chapter.
Check Also: AP EAMCET Syllabus 2025
AP EAMCET Probability Weightage of Previous Years
Here’s a table showing the year-wise weightage of the Probability chapter in AP EAMCET from 2019 to 2024.
Year | Weightage of Probability Chapter (%) | Total Marks (Out of 80) |
2019 | 3.75% | 3-4 marks |
2020 | 3.75% | 3-4 marks |
2021 | 3.75% | 3-4 marks |
2022 | 3.75% | 3-4 marks |
2023 | 3.75% | 3-4 marks |
2024 | 3.75% | 3-4 marks |
The Probability chapter has consistently held a weightage of around 3.75% in the AP EAMCET exam from 2019 to 2024.
Important Topics of AP EAMCET 2025 Probability and Weightage
Here’s a table outlining the important Probability topics for AP EAMCET 2025, along with their expected weightage:
Important Topics in Probability | Significance in AP EAMCET 2025 | Expected Weightage (%) |
Basic Probability Concepts | Fundamental understanding; frequently asked direct questions. | 1% - 2% |
Conditional Probability | Key topic; often tested with problems involving dice, cards, or events. | 1% |
Bayes' Theorem | Advanced concept; occasionally featured in more complex problems. | 0.5% - 1% |
Permutations and Combinations | Essential for selection-based problems; supports probability calculations. | 1% |
Independent and Dependent Events | Commonly tested; focuses on determining if events affect each other’s outcome. | 0.5% - 1% |
Random Variables and Distributions | Related to expected values; requires knowledge of discrete and continuous variables. | 0.5% - 1% |
Binomial Probability Distribution | Frequently appears in problems involving trials and successes. | 0.5% - 1% |
Sample Probability Questions for AP EAMCET 2025
Here are some sample questions to illustrate the type of Probability problems you might encounter in AP EAMCET:
Question 1:
A bag contains 4 red balls, 3 blue balls, and 5 green balls. If two balls are drawn randomly, what is the probability that both balls are red?
- Solution: Total number of balls = 4 + 3 + 5 = 12 Probability of drawing two red balls: P(two red balls)=(42)(122)=666=111P(\text{two red balls}) = \frac{\binom{4}{2}}{\binom{12}{2}} = \frac{6}{66} = \frac{1}{11}P(two red balls)=(212)(24)=666=111
Question 2:
A die is thrown twice. What is the probability that the sum of the numbers rolled is 8?
- Solution: Total possible outcomes = 6 × 6 = 36 Favorable outcomes: (2,6), (3,5), (4,4), (5,3), (6,2) = 5 outcomes. Probability: P(sum is 8)=536P(\text{sum is 8}) = \frac{5}{36}P(sum is 8)=365
Question 3:
What is the probability of getting at least one head when three coins are tossed?
- Solution: Total outcomes = 23=82^3 = 823=8 (HHH, HHT, HTH, HTT, THH, THT, TTH, TTT) Probability of no head = 1 outcome (TTT). Probability of at least one head = 1−18=781 - \frac{1}{8} = \frac{7}{8}1−81=87.
Question 4:
What is the probability of getting exactly two heads when three coins are tossed?
-
Solution: Total outcomes = 23=82^3 = 823=8. Outcomes with exactly two heads are (HHT, HTH, and THH), so there are 3 favourable outcomes.
Probability = 38\frac{3}{8}83.
Question 5:
A die is rolled. What is the probability of rolling a number greater than 4?
-
Solution: Possible outcomes greater than 4 are 5 and 6, so there are 2 favorable outcomes out of 6.
Probability = 26=13\frac{2}{6} = \frac{1}{3}62=31.
Question 6:
A bag contains 5 red balls and 3 blue balls. If one ball is drawn at random, what is the probability it is red?
-
Solution: Total balls = 5 + 3 = 8. Favorable outcomes (red) = 5.
Probability = 58\frac{5}{8}85.
Question 7:
Two dice are rolled together. What is the probability that the sum of the numbers is 7?
-
Solution: Possible outcomes for a sum of 7 are (1,6), (2,5), (3,4), (4,3), (5,2), (6,1), totalling 6 favourable outcomes out of 36.
Probability = 636=16\frac{6}{36} = \frac{1}{6}366=61.
In conclusion, Probability is a high-scoring and important topic in the Mathematics section of AP EAMCET 2025. With an approximate weightage of 3.75%, mastering this topic can significantly boost your overall score. By focusing on understanding the fundamental concepts, practising regularly, and revising frequently, you can ensure that you are well-prepared for the Probability questions in AP EAMCET 2025.
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