Download Probability Case Study PDF for Class 12 with Solutions

Probability case study class 12 pdf download

Probability case study class 12 pdf download helps students practice application-based questions aligned with the CBSE exam pattern. Moreover, these questions improve logical thinking and accuracy. Therefore, students gain confidence in handling real-life scenarios. As a result, performance in board exams improves significantly.

Importance of Probability Case Study Questions

Case study questions from Probability focus on real-world situations and data interpretation. Additionally, they strengthen conceptual clarity. However, regular practice is essential for scoring well. Consequently, students become familiar with multi-step problem-solving.

How to Prepare Using Case Study PDFs

Students should start by understanding basic probability formulas. Then, they must solve case study questions daily. Furthermore, revising solved examples helps avoid common mistakes. Ultimately, consistent practice leads to better exam results.

Case Study 5: Probability case study class 12 pdf download

Case Study Description: In a Grade 12 class, student performance in Mathematics and Physics is being analyzed. The school principal considers passing in Mathematics ($M$) and passing in Physics ($P$) as **independent events** for any randomly chosen student. The probability that a student passes in **Mathematics** is $P(M) = 0.80$, and the probability that a student passes in **Physics** is $P(P) = 0.70$.

The principal wants to calculate various probabilities related to a student’s performance in both subjects. Since the events are independent, the **Multiplication Theorem of Probability** for independent events simplifies the calculation of the probability of passing both or failing both. This assumption of independence means that a student’s preparation or aptitude in one subject does not influence their outcome in the other subject. While this is often a simplification in real life, it serves as a powerful model for understanding the basic rules of probability. The goal is to calculate the probabilities of compound events, such as passing at least one subject, and failing both, and to test the conceptual understanding of conditional probability for independent events.

Theory and Formulae Related to Independent Events:

  • **Joint Probability (Independent Events)**: $P(A \cap B) = P(A) \cdot P(B)$
  • **Union of Events**: $P(A \cup B) = P(A) + P(B) – P(A \cap B)$
  • **Conditional Probability (Independent Events)**: $P(A|B) = P(A)$
  • **Complementary Events**: $P(A’) = 1 – P(A)$

1. What is the probability that a randomly selected student passes in both Mathematics and Physics? $P(M \cap P)$

Solution: Since $M$ and $P$ are independent: $$P(M \cap P) = P(M) \cdot P(P) = 0.80 \times 0.70 = 0.56$$ Correct answer is option **(a)**.

2. What is the probability that a randomly selected student fails in both Mathematics and Physics? $P(M’ \cap P’)$

Solution: $P(M’) = 1 – 0.80 = 0.20$. $P(P’) = 1 – 0.70 = 0.30$. Since $M’$ and $P’$ are independent: $$P(M’ \cap P’) = P(M’) \cdot P(P’) = 0.20 \times 0.30 = 0.06$$ Correct answer is option **(b)**.

3. What is the probability that a randomly selected student passes in at least one of the two subjects? $P(M \cup P)$

Solution: Using the complementary event (failing both, from Q2): $$P(M \cup P) = 1 – P(M’ \cap P’) = 1 – 0.06 = 0.94$$ Alternatively: $P(M \cup P) = 0.80 + 0.70 – 0.56 = 0.94$. Correct answer is option **(c)**.

4. What is the probability that the student passes in Mathematics, given that they have passed in Physics? $P(M|P)$

Solution: Since $M$ and $P$ are **independent events**, the conditional probability $P(M|P)$ is simply the marginal probability $P(M)$. $$P(M|P) = P(M) = 0.80$$ Correct answer is option **(c)**.

5. Let $X$ be the random variable representing the number of subjects passed (out of 2). What is the probability $P(X=1)$, that a student passes in exactly one subject?

Solution: $P(X=1)$ is the sum of two mutually exclusive outcomes: 1. Pass Math, Fail Physics: $P(M \cap P’) = P(M) \cdot P(P’) = 0.80 \times 0.30 = 0.24$ 2. Fail Math, Pass Physics: $P(M’ \cap P) = P(M’) \cdot P(P) = 0.20 \times 0.70 = 0.14$ $$P(X=1) = 0.24 + 0.14$$ Correct answer is option **(a)**.

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