Probability case study class 12 with solutions

Probability Case Studies for Class 12 Explained with Step-by-Step Solutions

Probability Case Study Class 12 with Solutions

Probability case study class 12 with solutions helps students understand real-life applications of probability concepts. Moreover, these questions improve analytical thinking and exam readiness. Students learn how to calculate probabilities using logical steps and formulas. Therefore, regular practice builds confidence and accuracy. However, understanding the question context is equally important.

Understanding Case Study Based Questions

Case study questions present situations based on everyday events. For example, students may analyze cards, dice, or data-based scenarios. As a result, learners apply theoretical concepts practically. In addition, these questions test interpretation skills and formula application. Thus, consistent practice improves performance.

Preparation Tips for Class 12 Probability

Firstly, revise probability formulas thoroughly. Then, practice solved examples regularly. Meanwhile, focus on step-wise solutions to avoid errors. Additionally, reviewing previous year questions enhances exam strategy. Consequently, students gain better control over complex case study problems.

Case Study 1: Probability Case Studies for Class 12 Explained with Step-by-Step Solutions

Case Study Description: A company produces \textbf{LED bulbs} in two different plants, \textbf{Plant I} and \textbf{Plant II}. Plant I manufactures $60\%$ of the total bulbs, and Plant II manufactures the remaining $40\%$. The quality control department has determined the probability of a bulb being \textbf{defective} based on the plant of origin. The probability that a bulb produced by Plant I is defective is $5\%$, and the probability that a bulb produced by Plant II is defective is $2\%$. The company packages all the bulbs together for sale in the market. A buyer, Ramesh, purchases a large box of these LED bulbs. He is concerned about the reliability and quality of the production process. He wants to use the principles of probability to analyze the production data. Ramesh decides to randomly select one bulb from the box to test its quality. He is interested in finding the probability that the selected bulb is defective. Furthermore, if he finds a selected bulb to be defective, he wants to determine the probability that it came from a specific plant, say Plant I. This is a classic example where the concepts of \textbf{conditional probability}, \textbf{multiplication theorem of probability}, and \textbf{Bayes’ Theorem} are essential for understanding the overall quality and the source of potential defects. The relative contribution of each plant to the total production and their individual defect rates influence the final probability of finding a defective item. Let $E_1$ be the event that the bulb is produced by \textbf{Plant I}, and $E_2$ be the event that the bulb is produced by \textbf{Plant II}. Let $A$ be the event that the selected bulb is \textbf{defective}. We are given the following probabilities: $P(E_1) = 0.60$, $P(E_2) = 0.40$, $P(A|E_1) = 0.05$, and $P(A|E_2) = 0.02$. Ramesh needs to calculate the total probability $P(A)$ and the posterior probability $P(E_1|A)$.

Theory and Formulae Related to Probability:

  • Multiplication Theorem (Joint Probability): $P(E \cap A) = P(E) \cdot P(A|E)$
  • Theorem of Total Probability: $P(A) = P(E_1) \cdot P(A|E_1) + P(E_2) \cdot P(A|E_2)$
  • Bayes’ Theorem: $P(E_1|A) = \frac{P(E_1 \cap A)}{P(A)} = \frac{P(E_1) \cdot P(A|E_1)}{P(A)}$
  • Complementary Events: $P(A’) = 1 – P(A)$

1. What is the probability that a randomly selected bulb from the box is produced by Plant II and is defective?

Solution: This uses the Multiplication Theorem of Probability: $$P(E_2 \cap A) = P(E_2) \cdot P(A|E_2) = 0.40 \times 0.02 = 0.008$$ Correct answer is option **(b)**.

2. What is the probability that a randomly selected bulb from the box is defective? (Total Probability)

Solution: This uses the Theorem of Total Probability: $$P(A) = P(E_1) \cdot P(A|E_1) + P(E_2) \cdot P(A|E_2)$$ $$P(A) = (0.60 \times 0.05) + (0.40 \times 0.02) = 0.030 + 0.008 = 0.038$$ Correct answer is option **(d)**. (Note: Option (d) has been corrected from $0.032$ to $0.038$ to match the correct calculation.)

3. If a randomly selected bulb is found to be defective, what is the probability that it was produced by Plant I? (Bayes’ Theorem)

Solution: This uses Bayes’ Theorem: $$P(E_1|A) = \frac{P(E_1 \cap A)}{P(A)} = \frac{P(E_1) \cdot P(A|E_1)}{P(A)}$$ Using results from Q1 and Q2: $$P(E_1|A) = \frac{0.030}{0.038} = \frac{30}{38}$$ Correct answer is option **(a)**.

4. What is the conditional probability of a bulb being non-defective, given that it was produced in Plant II?

Solution: Let $A’$ be non-defective. $P(A’|E_2) = 1 – P(A|E_2)$ (Complementary events within the same plant). $$P(A’|E_2) = 1 – 0.02 = 0.98$$ Correct answer is option **(c)**.

5. Let $X$ be the random variable representing the number of defective bulbs when 2 bulbs are randomly selected from the entire lot (with replacement). If $P(\text{Defective}) = 0.038$, what is the probability $P(X=0)$?

Solution: Let $p = P(\text{Defective}) = 0.038$ and $q = P(\text{Non-defective}) = 1 – p = 0.962$. $P(X=0)$ means 0 defective bulbs in 2 trials (both are non-defective). Since selection is with replacement (independent): $$P(X=0) = q \cdot q = q^2 = (0.962)^2$$ Correct answer is option **(b)**.

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